17 câu Trắc nghiệm Toán 8 Bài 6: Phép cộng, phép trừ các phân thức đại số có đáp án (Vận dụng)

=$\frac{6{\mathrm{x}}^2+8\mathrm{x}+7+{\mathrm{x}}^2-\mathrm{x}-6{\mathrm{x}}^2-6\mathrm{x}-6}{(\mathrm{x}-1\left)\right({\mathrm{x}}^2+\mathrm{x}+1)}$

=$\frac{{\mathrm{x}}^2+\mathrm{x}+1}{(\mathrm{x}-1\left)\right({\mathrm{x}}^2+\mathrm{x}+1)}=\frac{1}{\mathrm{x}-1}$

Thay x = $\frac{1}{2}$ vào P =$\frac{1}{\mathrm{x}-1}$ ta được P = $\frac{1}{\frac{1}{2}-1}=\frac{1}{\frac{-1}{2}}=-2$.

Đáp án cần chọn là: A

Ta có B = $\frac{\mathrm{x}}{{\mathrm{x}}^3+1}+\frac{1-\mathrm{x}}{{\mathrm{x}}^2-\mathrm{x}+1}+\frac{1}{\mathrm{x}+1}$

=$\frac{\mathrm{x}}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}+\frac{(1-\mathrm{x}\left)\right(\mathrm{x}+1)}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}+\frac{1.({\mathrm{x}}^2-\mathrm{x}+1)}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}$

=$\frac{\mathrm{x}+1-{\mathrm{x}}^2+{\mathrm{x}}^2-\mathrm{x}+1}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}=\frac{2}{{\mathrm{x}}^3+1}$

Thay x = -2 vào B =$\frac{2}{{\mathrm{x}}^3+1}$  ta được B =$\frac{2}{{(-2)}^3+1}=\frac{-2}{7}$

Đáp án cần chọn là: C

Ta có

M = $\frac{10}{(\mathrm{x}+2\left)\right(3-\mathrm{x})}-\frac{12}{(3-\mathrm{x}\left)\right(3+\mathrm{x})}-\frac{1}{(\mathrm{x}+3\left)\right(\mathrm{x}+2)}$

=$\frac{10(\mathrm{x}+3)-12(\mathrm{x}+2)-(3-\mathrm{x})}{(\mathrm{x}+2\left)\right(3-\mathrm{x}\left)\right(\mathrm{x}+3)}$

=$\frac{10\mathrm{x}+30-12\mathrm{x}-24-3+\mathrm{x}}{(\mathrm{x}+2\left)\right(3-\mathrm{x}\left)\right(\mathrm{x}+3)}$

=$\frac{-\mathrm{x}+3}{(\mathrm{x}+2\left)\right(3-\mathrm{x}\left)\right(\mathrm{x}+3)}=\frac{1}{(\mathrm{x}+2\left)\right(\mathrm{x}+3)}$

Thay x = -0,25 vào M =$\frac{1}{(\mathrm{x}+2\left)\right(\mathrm{x}+3)}$  ta được

M=$\frac{1}{(-0,25+2\left)\right(-0,25+3)}=\frac{16}{77}$

Đáp án cần chọn là: D

Ta có

M =$\frac{10}{(\mathrm{x}+2\left)\right(3-\mathrm{x})}-\frac{12}{(3-\mathrm{x}\left)\right(3+\mathrm{x})}-\frac{1}{(\mathrm{x}+3\left)\right(\mathrm{x}+2)}$

=$\frac{10(\mathrm{x}+3)-12(\mathrm{x}+2)-(3-\mathrm{x})}{(\mathrm{x}+2\left)\right(3-\mathrm{x}\left)\right(\mathrm{x}+3)}$

=$\frac{10\mathrm{x}+30-12\mathrm{x}-24-3+\mathrm{x}}{(\mathrm{x}+2\left)\right(3-\mathrm{x}\left)\right(\mathrm{x}+3)}$

=$\frac{-\mathrm{x}+3}{(\mathrm{x}+2\left)\right(3-\mathrm{x}\left)\right(\mathrm{x}+3)}=\frac{1}{(\mathrm{x}+2\left)\right(\mathrm{x}+3)}$

Thay x = -0,25 vào M = $\frac{1}{(\mathrm{x}+2\left)\right(\mathrm{x}+3)}$ ta được

M=$\frac{1}{(-0,25+2\left)\right(-0,25+3)}=\frac{16}{77}$

Đáp án cần chọn là: D

Đáp án cần chọn là: D

Ta có

Đáp án cần chọn là: C

Ta có

$\begin{array}{l}\frac{1}{1-\mathrm{x}}+\frac{1}{1+\mathrm{x}}+\frac{2}{1+{\mathrm{x}}^2}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\\ =\frac{1+\mathrm{x}+1-\mathrm{x}}{(1-\mathrm{x}\left)\right(1+\mathrm{x})}+\frac{2}{1+{\mathrm{x}}^2}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\end{array}$

$\begin{array}{l}=\frac{2}{1-{\mathrm{x}}^2}+\frac{2}{1+{\mathrm{x}}^2}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\\ =\frac{2(1+{\mathrm{x}}^2)+2(1-{\mathrm{x}}^2)}{(1-{\mathrm{x}}^2\left)\right(1+{\mathrm{x}}^2)}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\end{array}$

$\frac{4}{1-{\mathrm{x}}^4}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}=\frac{4(1+{\mathrm{x}}^4)+4(1-{\mathrm{x}}^4)}{(1-{\mathrm{x}}^4\left)\right(1+{\mathrm{x}}^4)}+\frac{8}{1+{\mathrm{x}}^8}$

$\frac{8}{1-{\mathrm{x}}^8}+\frac{8}{1+{\mathrm{x}}^8}=\frac{8(1+{\mathrm{x}}^8)+8(1-{\mathrm{x}}^8)}{(1-{\mathrm{x}}^8\left)\right(1+{\mathrm{x}}^8)}=\frac{16}{1-{\mathrm{x}}^{16}}$

Vậy số cần điền là 16.

Đáp án cần chọn là: A

Ta có: $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}(\mathrm{x}+1)}+...+\frac{1}{(\mathrm{x}+9\left)\right(\mathrm{x}+10)}$

=$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}+\frac{1}{\mathrm{x}+1}-\frac{1}{\mathrm{x}+2}...+\frac{1}{\mathrm{x}+9}-\frac{1}{\mathrm{x}+10}$

=$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}}+0+...+0-\frac{1}{\mathrm{x}+10}$

=$\frac{2}{\mathrm{x}}-\frac{1}{\mathrm{x}+10}$

=$\frac{2\mathrm{x}+20-\mathrm{x}}{\mathrm{x}(\mathrm{x}+10)}=\frac{\mathrm{x}+20}{\mathrm{x}(\mathrm{x}+10)}$

Đáp án cần chọn là: A

ĐK: x ≠ 1.

$\frac{\mathrm{x}-1}{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{P}=\frac{2}{\mathrm{x}-1}+\frac{3\mathrm{x}}{1-{\mathrm{x}}^3}$

P=$\frac{\mathrm{x}-1}{{\mathrm{x}}^2+\mathrm{x}+1}-\frac{2}{\mathrm{x}-1}+\frac{3\mathrm{x}}{{\mathrm{x}}^3-1}$

P=$\frac{{(\mathrm{x}-1)}^2-2({\mathrm{x}}^2+\mathrm{x}+1)+3\mathrm{x}}{(\mathrm{x}-1\left)\right({\mathrm{x}}^2+\mathrm{x}+1)}$

P=$\frac{{\mathrm{x}}^2-2\mathrm{x}+1-2{\mathrm{x}}^2-2\mathrm{x}-2+3\mathrm{x}}{(\mathrm{x}-1\left)\right({\mathrm{x}}^2+\mathrm{x}+1)}$

P=$\frac{-{\mathrm{x}}^2-\mathrm{x}-1}{(\mathrm{x}-1\left)\right({\mathrm{x}}^2+\mathrm{x}+1)}$

P=$-\frac{1}{\mathrm{x}-1}$

Đáp án cần chọn là: D

ĐK: x ≠ {-2; 2; 3}.

P +$\frac{4\mathrm{x}-12}{{\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}+12}=\frac{3}{\mathrm{x}-3}-\frac{{\mathrm{x}}^2}{4-{\mathrm{x}}^2}$

P=$\frac{3}{\mathrm{x}-3}-\frac{{\mathrm{x}}^2}{4-{\mathrm{x}}^2}-\frac{4\mathrm{x}-12}{{\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}+12}$

P=$\frac{3}{\mathrm{x}-3}+\frac{{\mathrm{x}}^2}{(\mathrm{x}-2)(\mathrm{x}+2)}-\frac{4\mathrm{x}-12}{{\mathrm{x}}^2(\mathrm{x}-3)-4(\mathrm{x}-3)}$

P=$\frac{3({\mathrm{x}}^2-4)}{(\mathrm{x}-3\left)\right({\mathrm{x}}^2-4)}+\frac{{\mathrm{x}}^2(\mathrm{x}-3)}{(\mathrm{x}-3)({\mathrm{x}}^2-4)}-\frac{4\mathrm{x}-12}{(\mathrm{x}-3)({\mathrm{x}}^2-4)}$

P=$\frac{3{\mathrm{x}}^2-12+{\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}+12}{(\mathrm{x}-3\left)\right({\mathrm{x}}^2-4)}$

P=$\frac{{\mathrm{x}}^3-4\mathrm{x}}{(\mathrm{x}-3\left)\right(\mathrm{x}-2\left)\right(\mathrm{x}+2)}$

P=$\frac{\mathrm{x}({\mathrm{x}}^2-4)}{(\mathrm{x}-3\left)\right(\mathrm{x}-2\left)\right(\mathrm{x}+2)}=\frac{\mathrm{x}}{\mathrm{x}-3}$

Đáp án cần chọn là: B

Ta có

$\frac{\mathrm{x}}{1-{\mathrm{x}}^2}+\frac{\mathrm{y}}{1-{\mathrm{y}}^2}+\frac{\mathrm{z}}{1-{\mathrm{z}}^2}$

=$\frac{\mathrm{x}(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)+\mathrm{y}(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{z}}^2)+\mathrm{z}(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2)}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

=$\frac{\mathrm{x}(1-{\mathrm{z}}^2-{\mathrm{y}}^2+{\mathrm{z}}^2{\mathrm{y}}^2)+\mathrm{y}(1-{\mathrm{x}}^2-{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{z}}^2)+\mathrm{z}(1-{\mathrm{x}}^2-{\mathrm{y}}^2+{\mathrm{x}}^2{\mathrm{y}}^2)}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

=$\frac{\mathrm{x}-{\mathrm{xz}}^2-{\mathrm{xy}}^2+{\mathrm{xy}}^2{\mathrm{z}}^2+\mathrm{y}-{\mathrm{x}}^2\mathrm{y}-{\mathrm{yz}}^2+{\mathrm{yz}}^2{\mathrm{x}}^2+\mathrm{z}-{\mathrm{zx}}^2-{\mathrm{zy}}^2+{\mathrm{zx}}^2{\mathrm{y}}^2}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

$=\frac{(\mathrm{x}-\mathrm{y}{\mathrm{x}}^2-{\mathrm{x}}^2\mathrm{z})+(\mathrm{y}-{\mathrm{xy}}^2-{\mathrm{zy}}^2)+(\mathrm{z}-{\mathrm{xz}}^2-{\mathrm{yz}}^2)+(\mathrm{x}{\mathrm{y}}^2{\mathrm{z}}^2+{\mathrm{yz}}^2{\mathrm{x}}^2+{\mathrm{zx}}^2{\mathrm{y}}^2)}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

=$\frac{\mathrm{x}(1-\mathrm{xy}-x\mathrm{z})+\mathrm{y}(1-\mathrm{xy}-\mathrm{yz})+\mathrm{z}(1-\mathrm{xz}-\mathrm{zy})+\mathrm{xyz}(\mathrm{yz}+\mathrm{xz}+\mathrm{xy})}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

=$\frac{\mathrm{x}.\mathrm{yz}+\mathrm{y}.\mathrm{xz}+\mathrm{z}.\mathrm{xy}+\mathrm{xyz}}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

=$\frac{4\mathrm{xyz}}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

Đáp án cần chọn là: C

Vì  $\frac{{\mathrm{x}}^2}{\mathrm{y}+\mathrm{z}}+\frac{{\mathrm{y}}^2}{\mathrm{x}+\mathrm{z}}+\frac{{\mathrm{z}}^2}{\mathrm{x}+\mathrm{y}}$= 0 nên ta có

x + y + z = x + y + z + 0

= x + y + z +$\frac{{\mathrm{x}}^2}{\mathrm{y}+\mathrm{z}}+\frac{{\mathrm{y}}^2}{\mathrm{x}+\mathrm{z}}+\frac{{\mathrm{z}}^2}{\mathrm{x}+\mathrm{y}}$

=$\left(\mathrm{x}+\frac{{\mathrm{x}}^2}{\mathrm{y}+\mathrm{z}}\right)+\left(\mathrm{y}+\frac{{\mathrm{y}}^2}{\mathrm{x}+\mathrm{z}}\right)+\left(\mathrm{z}+\frac{{\mathrm{z}}^2}{\mathrm{x}+\mathrm{y}}\right)$

=$\mathrm{x}\left(1+\frac{\mathrm{x}}{\mathrm{y}+\mathrm{z}}\right)+\mathrm{y}\left(1+\frac{\mathrm{y}}{\mathrm{x}+\mathrm{z}}\right)+\mathrm{z}\left(1+\frac{\mathrm{z}}{\mathrm{x}+\mathrm{y}}\right)$

=$\mathrm{x}\left(\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{y}+\mathrm{z}}\right)+\mathrm{y}\left(\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{x}+\mathrm{z}}\right)+\mathrm{z}\left(\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{x}+\mathrm{y}}\right)$

=$(\mathrm{x}+\mathrm{y}+\mathrm{z})\left(\frac{\mathrm{x}}{\mathrm{y}+\mathrm{z}}+\frac{\mathrm{y}}{\mathrm{x}+\mathrm{z}}+\frac{\mathrm{z}}{\mathrm{x}+\mathrm{y}}\right)$

Suy ra x + y + z =$(\mathrm{x}+\mathrm{y}+\mathrm{z})\left(\frac{\mathrm{x}}{\mathrm{y}+\mathrm{z}}+\frac{\mathrm{y}}{\mathrm{x}+\mathrm{z}}+\frac{\mathrm{z}}{\mathrm{x}+\mathrm{y}}\right)$

Mà x + y + z ≠ 0 nên $\frac{\mathrm{x}}{\mathrm{y}+\mathrm{z}}+\frac{\mathrm{y}}{\mathrm{x}+\mathrm{z}}+\frac{\mathrm{z}}{\mathrm{x}+\mathrm{y}}$ = 1

Hay A = 1.

Đáp án cần chọn là: D