16 câu hỏi
\[\frac{A}{B} - \frac{C}{D} = \frac{{A - C}}{{B - D}}\]
\[\frac{A}{B} - \frac{C}{D} = \frac{{AD}}{{BC}}\]
\[\frac{A}{B} - \frac{C}{D} = \frac{{AD - BC}}{{BD}}\]
\[\frac{A}{B} - \frac{C}{D} = \frac{{A - C}}{{BD}}\]
Lời giải
Đáp án đúng là: C
Quy đồng mẫu thức \[\frac{{\rm{A}}}{{\rm{B}}}\] và \[\frac{{\rm{C}}}{{\rm{D}}}\], ta có:
\[\frac{A}{B} = \frac{{AD}}{{BD}};\,\,\,\frac{C}{D} = \frac{{BC}}{{BD}}\].
Do đó \[\frac{A}{B} - \frac{C}{D} = \frac{{AD}}{{BD}} - \frac{{BC}}{{BD}} = \frac{{AD - BC}}{{BD}}\].
\[\frac{{2{\rm{x}} + 1}}{{{\rm{x}} + 1}}\]
\[\frac{{1 - 2{\rm{x}}}}{{{\rm{x}} + 1}}\]
\[\frac{{{\rm{x}} + 1}}{{2{\rm{x}} - 1}}\]
\[\frac{{{\rm{x}} + 1}}{{1 - 2{\rm{x}}}}\]
Lời giải
Đáp án đúng là: B
Phân thức đối của phân thức\[\frac{{2{\rm{x}} - 1}}{{{\rm{x}} + 1}}\] là \[ - \frac{{2x - 1}}{{x{\rm{ }} + {\rm{ }}1}} = \frac{{1 - 2x}}{{x{\rm{ }} + {\rm{ }}1}}\].
x + 2
2x
x
x – 2
Lời giải
Đáp án đúng là: D
\[\frac{{{{\rm{x}}^2}}}{{{\rm{x}} + 2}} - \frac{4}{{{\rm{x}} + 2}} = \frac{{{{\rm{x}}^2} - 4}}{{{\rm{x}} + 2}} = \frac{{\left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} + 2} \right)}}{{{\rm{x}} + 2}}\]
\[ = \frac{{\left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} + 2} \right):\left( {{\rm{x}} + 2} \right)}}{{\left( {{\rm{x}} + 2} \right):\left( {{\rm{x}} + 2} \right)}} = \frac{{{\rm{x}} - 2}}{1} = {\rm{x}} - 2\].
\[\frac{2}{{{\rm{x}} - 2}}\]
\[\frac{2}{{2 - {\rm{x}}}}\]
\[\frac{1}{{\rm{x}}}\]
\[\frac{1}{{{\rm{x}} + 2}}\]
Lời giải
Đáp án đúng là: B
\[\frac{{x - 1}}{{{x^2} - 2x}} + A = \frac{{ - x - 1}}{{{x^2} - 2x}}\]
Suy ra \[A = \frac{{ - x - 1}}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x}}}} - \frac{{{\rm{x}} - {\rm{1}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x}}}}\]
\[ = \frac{{ - x - 1 - \left( {x - 1} \right)}}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x}}}}\]\[ = \frac{{ - x - 1 - x + 1}}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x}}}}\]
\[ = \frac{{ - 2x}}{{{{\rm{x}}^{\rm{2}}} - {\rm{2x}}}} = \frac{{ - 2x}}{{x\left( {x - 2} \right)}}\]\[ = \frac{{ - 2}}{{x - 2}} = \frac{2}{{2 - x}}\].
\[\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}} - {\rm{7x}} - {\rm{2}}}}{{{\rm{12x}}\left( {{\rm{x}} - {\rm{1}}} \right)\left( {{\rm{x + 1}}} \right)}}\]
\[4{{\rm{x}}^2} - 7{\rm{x}} + 2\]
\[4{{\rm{x}}^2} - 7{\rm{x}} - 2\]
\[{\rm{12x}}\left( {{\rm{x}} - {\rm{1}}} \right)\left( {{\rm{x + 1}}} \right)\]
Lời giải
Đáp án đúng là: C
\[A = \frac{{2x - 1}}{{6{x^2} - 6x}} - \frac{3}{{4{x^2} - 4}}\]\[ = \frac{{2x - 1}}{{6x\left( {x - 1} \right)}} - \frac{3}{{4\left( {{x^2} - 1} \right)}}\]
\[ = \frac{{2x - 1}}{{6x\left( {x - 1} \right)}} - \frac{3}{{4\left( {x - 1} \right)\left( {x + 1} \right)}}\]\[ = \frac{{2\left( {2x - 1} \right)\left( {x + 1} \right) - 3.3x}}{{12x\left( {x - 1} \right)\left( {x + 1} \right)}}\]
\[ = \frac{{2\left( {2{x^2} - x + 2x - 1} \right) - 9x}}{{12x\left( {x - 1} \right)\left( {x + 1} \right)}}\]\[ = \frac{{2\left( {2{x^2} + x - 1} \right) - 9x}}{{12x\left( {x - 1} \right)\left( {x + 1} \right)}}\]
\[ = \frac{{4{x^2} + 2x - 2 - 9x}}{{12x\left( {x - 1} \right)\left( {x + 1} \right)}}\]\[ = \frac{{4{x^2} - 7x - 2}}{{12x\left( {x - 1} \right)\left( {x + 1} \right)}}\].\[{\rm{B}} = \frac{1}{{2\,\,020}}\]
\[{\rm{B}} = \frac{1}{{202\,\,000}}\]
\[{\rm{B}} = \frac{1}{{200\,\,200}}\]
\[{\rm{B}} = \frac{1}{{20\,\,200}}\]
Lời giải
Đáp án đúng là: B
\[{\rm{B}} = \frac{1}{{{\rm{x}} - 23}} - \frac{1}{{{\rm{x}} - 3}} = \frac{{x - 3}}{{\left( {x - 23} \right)\left( {x - 3} \right)}} - \frac{{x - 23}}{{\left( {x - 23} \right)\left( {x - 3} \right)}}\]
\[ = \frac{{\left( {x - 3} \right) - \left( {x - 23} \right)}}{{\left( {x - 23} \right)\left( {x - 3} \right)}} = \frac{{x - 3 - x + 23}}{{\left( {x - 23} \right)\left( {x - 3} \right)}} = \frac{{20}}{{\left( {x - 23} \right)\left( {x - 3} \right)}}\]
Với x = 2023, ta có:
\[{\rm{B}} = \frac{{20}}{{\left( {2023 - 23} \right)\left( {2023 - 3} \right)}} = \frac{{20}}{{2000\,.\,2020}}\]
\[ = \frac{{20}}{{20\,.\,\,100\,.\,2020}} = \frac{1}{{100\,.\,2020}} = \frac{1}{{202\,\,000}}\].
x = 0
\(x = \frac{1}{2}\)
x = 1
\[{\rm{x}} = \frac{3}{2}\]
Lời giải
Đáp án đúng là: D
Ta có \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}}\]\[ = \frac{2}{{{\rm{x}} + 3}} + \frac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]
\[ = \frac{{2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \frac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]
\[ = \frac{{2\left( {x - 3} \right) + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x - 6 + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]
Mà \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}} = 0\] nên \[\frac{{2x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\]
\[2x - 3 = 0\]
\[2x = 3\]
\[x = \frac{3}{2}\]
Vậy \[x = \frac{3}{2}\].
Học thông minh hơn mỗi ngày
Xem toàn bộ 16 câu hỏi và nhiều hơn nữa với tài khoản miễn phí
\[A = \frac{{ - 6{x^2} + {\rm{ }}2x - 15}}{{\left( {x - 1} \right)\left( {{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1} \right)}}\]
\[A = \frac{{6{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1} \right)}}\]
\[A = \frac{{6{x^2} + {\rm{ }}15}}{{\left( {x - 1} \right)\left( {{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1} \right)}}\]
\[A = \frac{{ - 6{x^2} - 15}}{{\left( {x - 1} \right)\left( {{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1} \right)}}\]
Lời giải
Đáp án đúng là: D
\[A = \frac{{2{x^2} + {\rm{ }}x - 3}}{{{x^3} - 1}} - \frac{{x - 5}}{{{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1}} - \frac{7}{{x - 1}}\]
\[ = \frac{{2{x^2} + x - 3}}{{{{\rm{x}}^{\rm{3}}} - {\rm{1}}}} - \left( {\frac{{x - 5}}{{{x^2} + x + 1}} + \frac{7}{{x - 1}}} \right)\]
\[ = \frac{{2{x^2} + x - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \left[ {\frac{{\left( {x - 5} \right)\left( {x - 1} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}} + \frac{{7\left( {{x^2} + x + 1} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}}} \right]\]
\[ = \frac{{2{x^2} + x - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \left[ {\frac{{{x^2} - 5x - x + 5}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}} + \frac{{7{x^2} + 7x + 7}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}}} \right]\]
\[ = \frac{{2{x^2} + x - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \frac{{{x^2} - 5x - x + 5 + 7{x^2} + 7x + 7}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
\[ = \frac{{2{x^2} + x - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \frac{{8{x^2} + x + 12}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
\[ = \frac{{\left( {2{x^2} + x - 3} \right) - \left( {8{x^2} + x + 12} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
\[ = \frac{{2{x^2} + x - 3 - 8{x^2} - x - 12}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{ - 6{x^2} - 15}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
\[A = \frac{{11}}{2}\]
\[A = \frac{{13}}{2}\]
\[{\rm{A}} = \frac{{15}}{2}\]
\[{\rm{A}} = \frac{{17}}{2}\]
Lời giải
Đáp án đúng là: D
\[A = \frac{5}{{2x}} + \frac{{2x - 3}}{{2x - 1}} + \frac{{4{x^2} + {\rm{ }}3}}{{8{x^2} - 4x}}\]
\[ = \frac{5}{{2x}} + \frac{{2x - 3}}{{2x - 1}} + \frac{{4{x^2}}}{{4x\left( {2x - 1} \right)}}\]
\[ = \frac{{5\,.\,2\left( {2x - 1} \right)}}{{4x\left( {2x - 1} \right)}} + \frac{{4x\left( {2x - 3} \right)}}{{4x\left( {2x - 1} \right)}} + \frac{{4{x^2} + 3}}{{4x\left( {2x - 1} \right)}}\]
\[ = \frac{{20x - 10}}{{4x\left( {2x - 1} \right)}} + \frac{{8{x^2} - 12x}}{{4x\left( {2x - 1} \right)}} + \frac{{4{x^2} + 3}}{{4x\left( {2x - 1} \right)}}\]
\[ = \frac{{20x - 10 + 8{x^2} - 12x + 4{x^2} + 3}}{{4x\left( {2x - 1} \right)}} = \frac{{12{x^2} + 8x - 7}}{{4x\left( {2x - 1} \right)}}\]
\[ = \frac{{12{x^2} - 6x + 14x - 7}}{{4x\left( {2x - 1} \right)}} = \frac{{6x\left( {2x - 1} \right) + 7\left( {2x - 1} \right)}}{{4x\left( {2x - 1} \right)}}\]
\[ = \frac{{\left( {6x + 7} \right)\left( {2x - 1} \right)}}{{4x\left( {2x - 1} \right)}} = \frac{{6x + 7}}{{4x}}\].
Với \[{\rm{x}} = \frac{1}{4}\], ta có:
\[{\rm{A}} = \frac{{6 \cdot \frac{1}{4} + 7}}{{4 \cdot \frac{1}{4}}} = \frac{{\frac{3}{2} + 7}}{1} = \frac{3}{2} + 7 = \frac{3}{2} + \frac{{14}}{2} = \frac{{17}}{2}\].A = 1
A = 0
\[{\rm{A}} = \frac{1}{2}\]
\[{\rm{A}} = \frac{{99}}{{100}}\]
Lời giải
Đáp án đúng là: D
\[{\rm{A}} = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{99.100}}\]
\[ = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ... + \left( {\frac{1}{{99}} - \frac{1}{{100}}} \right)\]
\[ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}}\]
\[ = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}\]
1
2
3
4
Lời giải
Đáp án đúng là: D
Ta có 3y – x = 6 nên x = 3y – 63
Thay x = 3y – 6 vào \[{\rm{A = }}\frac{{\rm{x}}}{{{\rm{y}} - {\rm{2}}}}{\rm{ + }}\frac{{{\rm{2x}} - {\rm{3y}}}}{{{\rm{x}} - {\rm{6}}}}\], ta được:
\[A = \frac{{3y - 6}}{{y - 2}} + \frac{{2\left( {3y - 6} \right) - 3y}}{{3y - 6 - 6}}\]
\[ = \frac{{3\left( {y - 2} \right)}}{{y - 2}} + \frac{{6y - 12 - 3y}}{{3y - 12}}\]
\[ = 3 + \frac{{3y - 12}}{{3y - 12}} = 3 + 1 = 4\]
0 < A < 1
A = 0
A = 1
\[{\rm{A}} = \frac{7}{4}\]
Lời giải
Đáp án đúng là: A
\[{\rm{A}} = \frac{{10}}{{\left( {{\rm{x}} + 2} \right)\left( {3 - {\rm{x}}} \right)}} - \frac{{12}}{{\left( {3 - {\rm{x}}} \right)\left( {3 + {\rm{x}}} \right)}} - \frac{1}{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} + 2} \right)}}\]
\[ = \frac{{10}}{{\left( {{\rm{x}} + 2} \right)\left( {3 - {\rm{x}}} \right)}} - \left[ {\frac{{12}}{{\left( {3 - x} \right)\left( {3 + x} \right)}} + \frac{1}{{\left( {x + 3} \right)\left( {x + 2} \right)}}} \right]\]
\[ = \frac{{10}}{{\left( {{\rm{x}} + 2} \right)\left( {3 - {\rm{x}}} \right)}} - \left[ {\frac{{12\left( {x + 2} \right) + \left( {3 - x} \right)}}{{\left( {3 - x} \right)\left( {x + 3} \right)\left( {x + 2} \right)}}} \right]\]
\[ = \frac{{10}}{{\left( {{\rm{x}} + 2} \right)\left( {3 - {\rm{x}}} \right)}} - \left[ {\frac{{12x + 24 + 3 - x}}{{\left( {3 - x} \right)\left( {x + 3} \right)\left( {x + 2} \right)}}} \right]\]
\[ = \frac{{10}}{{\left( {{\rm{x}} + 2} \right)\left( {3 - {\rm{x}}} \right)}} - \frac{{11x + 27}}{{\left( {3 - x} \right)\left( {x + 3} \right)\left( {x + 2} \right)}}\]
\[ = \frac{{10\left( {x + 3} \right)}}{{\left( {3 - x} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} - \frac{{11x + 27}}{{\left( {3 - x} \right)\left( {x + 2} \right)\left( {x + 3} \right)}}\]
\[ = \frac{{10\left( {x + 3} \right) - \left( {11x + 27} \right)}}{{\left( {3 - x} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} = \frac{{10x + 30 - 11x - 27}}{{\left( {3 - x} \right)\left( {x + 2} \right)\left( {x + 3} \right)}}\]
\[ = \frac{{ - x + 3}}{{\left( {3 - x} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} = \frac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\]
Tại \[{\rm{x}} = - \frac{3}{4}\]ta có\[{\rm{A}} = \frac{1}{{\left( {\frac{{ - 3}}{4} + 2} \right)\left( {\frac{{ - 3}}{4} + 3} \right)}} = \frac{1}{{\frac{5}{4} \cdot \frac{9}{4}}} = \frac{1}{{\frac{{45}}{{16}}}} = \frac{{16}}{{45}}\]
Vậy 0 < A < 1.
x = 0
x = 1
\(x = \pm 1\)
\[{\rm{x}} \in \left\{ {0;\,\,2} \right\}\]
Lời giải
Đáp án đúng là: D
\[{\rm{A = }}\frac{{{\rm{6}}{{\rm{x}}^{\rm{2}}}{\rm{ + 8x + 7}}}}{{{{\rm{x}}^{\rm{3}}} - {\rm{1}}}}{\rm{ + }}\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + x + 1}}}} - \frac{{\rm{6}}}{{{\rm{x}} - {\rm{1}}}}\]
\[ = \frac{{6{x^2} + 8x + 7}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \frac{x}{{{x^2} + x + 1}} - \frac{6}{{x - 1}}\]
\[ = \frac{{6{x^2} + 8x + 7 + x\left( {x - 1} \right) - 6\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
\[ = \frac{{6{x^2} + 8x + 7 + {x^2} - x - 6{x^2} - 6x - 6}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
\[ = \frac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{1}{{x - 1}}\]
Để \[{\rm{A}} \in \mathbb{Z}\] hay \[\frac{1}{{{\rm{x}} - 1}} \in \mathbb{Z}\] thì x – 1 ∈ Ư(1) = {−1; 1}.
Ta có bảng sau:
|
x – 1 |
−1 |
1 |
|
x |
0 (TM) |
2 (TM) |
1
2
3
4
Lời giải
Đáp án đúng là: C
Điều kiện:\(\left\{ {\begin{array}{*{20}{c}}{x - 3 \ne 0}\\{4 - {x^2} \ne 0}\\{{x^3} - 3{x^2} - 4x + 12 \ne 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \ne 3}\\{x \ne \pm 2}\end{array}} \right.\)
\[{\rm{A = }}\frac{{\rm{3}}}{{{\rm{x}} - {\rm{3}}}} - \frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{4}} - {{\rm{x}}^{\rm{2}}}}} - \frac{{{\rm{4x}} - {\rm{12}}}}{{{{\rm{x}}^{\rm{3}}} - {\rm{3}}{{\rm{x}}^{\rm{2}}} - {\rm{4x + 12}}}}\]
\[ = \frac{3}{{x - 3}} - \frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{4}} - {{\rm{x}}^{\rm{2}}}}} - \frac{{4x - 12}}{{{x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right)}}\]
\[ = \frac{3}{{x - 3}} + \frac{{{x^2}}}{{{x^2} - 4}} - \frac{{4x - 12}}{{\left( {{x^2} - 4} \right)\left( {x - 3} \right)}}\]
\[ = \frac{{3\left( {{x^2} - 4} \right) + {x^2}\left( {x - 3} \right) - \left( {4x - 12} \right)}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}}\]
\[ = \frac{{3{x^2} - 12 + {x^3} - 3{x^2} - 4x + 12}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}}\]
\[ = \frac{{{x^3} - 4x}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}} = \frac{{x\left( {{x^2} - 4} \right)}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}}\]
\[ = \frac{x}{{x - 3}} = 1 + \frac{3}{{x - 3}}\]
Để \[{\rm{A}} \in \mathbb{Z} \Rightarrow \frac{3}{{{\rm{x}} - 3}} \in \mathbb{Z} \Rightarrow \left( {{\rm{x}} - 3} \right) \in {\rm{U}}\left( 3 \right) = \left\{ { \pm \,1;\,\, \pm 3} \right\}\].
Ta có bảng sau:
|
x – 3 |
–3 |
–1 |
1 |
3 |
|
x |
0 (TM) |
2 (KTM) |
4 (TM) |
6 (TM) |
Vậy có 3 giá trị của x để biểu thức A có giá trị là một số nguyên.
\[\frac{1}{{2{\rm{x}}\left( {{\rm{x}} - 1} \right)}}\]
\[\frac{1}{{2{\rm{x}}\left( {{\rm{x}} + 1} \right)}}\]
\[\frac{2}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right)}}\]
\[\frac{{2{\rm{x}}}}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right)}}\]
Lời giải
Đáp án đúng là: A
\[{\rm{A}} = \frac{3}{{2{{\rm{x}}^2} + 2{\rm{x}}}} + \frac{{\left| {2{\rm{x}} - 1} \right|}}{{{{\rm{x}}^2} - 1}} - \frac{2}{{\rm{x}}}\]
\[ = \frac{3}{{2{{\rm{x}}^2} + 2{\rm{x}}}} + \frac{{2x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{2}{{\rm{x}}}\]
\[ = \frac{{3\left( {x - 1} \right) + 2x\left( {2x - 1} \right) - 4\left( {x - 1} \right)\left( {x + 1} \right)}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}}\]
\[ = \frac{{3x - 3 + 4{x^2} - 2x - 4{x^2} + 4}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}}\]
\[ = \frac{{x + 1}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{2x\left( {x - 1} \right)}}\]
16
8
4
20
Lời giải
Đáp án đúng là: A
\[\frac{1}{{1 - {\rm{x}}}} + \frac{1}{{1 + {\rm{x}}}} + \frac{2}{{1 + {{\rm{x}}^2}}} + \frac{4}{{1 + {{\rm{x}}^4}}} + \frac{8}{{1 + {{\rm{x}}^8}}}\]
\[ = \frac{{1 + x + 1 - x}}{{\left( {1 - x} \right)\left( {1 + x} \right)}} + \frac{2}{{1 + {{\rm{x}}^2}}} + \frac{4}{{1 + {{\rm{x}}^4}}} + \frac{8}{{1 + {{\rm{x}}^8}}}\]
\[ = \frac{2}{{1 - {{\rm{x}}^2}}} + \frac{2}{{1 + {{\rm{x}}^2}}} + \frac{4}{{1 + {{\rm{x}}^4}}} + \frac{8}{{1 + {{\rm{x}}^8}}}\]
\[ = \frac{{2\left( {1 + {{\rm{x}}^2}} \right) + 2\left( {1 - {{\rm{x}}^2}} \right)}}{{\left( {1 - {{\rm{x}}^2}} \right)\left( {1 + {{\rm{x}}^2}} \right)}} + \frac{4}{{1 + {{\rm{x}}^4}}} + \frac{8}{{1 + {{\rm{x}}^8}}}\]
\[ = \frac{4}{{1 - {{\rm{x}}^4}}} + \frac{4}{{1 + {{\rm{x}}^4}}} + \frac{8}{{1 + {{\rm{x}}^8}}}\]
\[ = \frac{{4\left( {1 + {{\rm{x}}^4}} \right) + 4\left( {1 - {{\rm{x}}^4}} \right)}}{{\left( {1 - {{\rm{x}}^4}} \right)\left( {1 + {{\rm{x}}^4}} \right)}} + \frac{8}{{1 + {{\rm{x}}^8}}}\]
\[ = \frac{8}{{1 - {{\rm{x}}^8}}} + \frac{8}{{1 + {{\rm{x}}^8}}} = \frac{{8\left( {1 + {{\rm{x}}^8}} \right) + 8\left( {1 - {{\rm{x}}^8}} \right)}}{{\left( {1 - {{\rm{x}}^8}} \right)\left( {1 + {{\rm{x}}^8}} \right)}} = \frac{{16}}{{1 - {x^{16}}}}\].
Bạn có thích hoạt động này không?
Ý kiến của bạn sẽ giúp chúng tôi đề xuất nội dung phù hợp hơn.
![[Năm 2022] Đề thi thử môn Vật lí THPT Quốc gia có lời giải - Đề số 21](/_next/image?url=https%3A%2F%2Fvideo.vietjack.com%2Fupload%2Fimages%2Fvj-image%2Fimage-3izficdefgfoel0s9j8s-1786772791.jpg&w=3840&q=75)
![[Năm 2022] Đề thi thử môn Vật lí THPT Quốc gia có lời giải - Đề số 20](/_next/image?url=https%3A%2F%2Fvideo.vietjack.com%2Fupload%2Fimages%2Fvj-image%2Fimage-emy0gtcksawvorajezvv-1786772460.jpg&w=3840&q=75)
