Hướng dẫn giải  

a) • Thay \[x = 0\,;{\rm{ }}y = - 2\] vào đa thức \[M\], ta được:

\[M = 23 \cdot {0^{23}} \cdot \left( { - 2} \right) - 22 \cdot 0 \cdot {\left( { - 2} \right)^{23}} + 21 \cdot \left( { - 2} \right) - 1\]\[ = 21 \cdot \left( { - 2} \right) - 1 = - 43\].

• Thay \[x = 0\,;{\rm{ }}y = - 2\] vào đa thức \[N,\] ta được:

\[N = - 22 \cdot 0 \cdot {\left( { - 2} \right)^{23}} - 42 \cdot \left( { - 2} \right) - 1\]\[ = - 42 \cdot \left( { - 2} \right) - 1 = 83\].

Vậy tại \(x = 0;\,y = - 2\) thì \[M = - 43\,;\,\,N = 83.\]

b) Ta có \(R = M - N = 23{x^{23}}y - 22x{y^{23}} + 21y - 1 + 22x{y^{23}} + 42y + 1\)

\[\, = 23{x^{23}}y + \left( { - 22x{y^{23}} + 22x{y^{23}}} \right) + \left( {21y + 42y} \right) + \left( { - 1 + 1} \right)\]\[ = 23{x^{23}}y + 63y\].

Vậy \[R = 23{x^{23}}y + 63y.\]

a) \(M + 5{x^2} - 2xy = 6{x^2} + 9xy - {y^2}\)

Suy ra \(M = 6{x^2} + 9xy - {y^2} - \left( {5{x^2} - 2xy} \right)\)

\( = 6{x^2} + 9xy - {y^2} - 5{x^2} + 2xy\)

\( = \left( {6{x^2} - 5{x^2}} \right) + \left( {9xy + 2xy} \right) - {y^2}\)

\( = {x^2} + 11xy - {y^2}\).

b) \(N - \left( {5{y^2} - xyz} \right) = xy + 2{y^2} - 3xyz + 4\)

Suy ra \(N = xy + 2{y^2} - 3xyz + 4 + \left( {5{y^2} - xyz} \right)\)

\( = xy + 2{y^2} - 3xyz + 4 + 5{y^2} - xyz\)

\[ = xy + \left( {2{y^2} + 5{y^2}} \right) + \left( { - 3xyz - xyz} \right) + 4\]

\[ = xy + 7{y^2} - 4xyz + 4\].

Hướng dẫn giải

Ta có \(\left( {2x + 3y} \right)\left( {{x^2} - xy + {y^2}} \right) - \left( {{x^3}{y^2} - {x^2}{y^3} + 4xy} \right):xy\)

\[ = 2{x^3} - 2{x^2}y + 2x{y^2} + 3{x^2}y - 3x{y^2} + 3{y^3} - \left( {{x^2}y - x{y^2} + 4} \right)\]

\[ = 2{x^3} - 2{x^2}y + 2x{y^2} + 3{x^2}y - 3x{y^2} + 3{y^3} - {x^2}y + x{y^2} - 4\]

\[ = 2{x^3} + \left( { - 2{x^2}y + 3{x^2}y - {x^2}y} \right) + \left( {2x{y^2} - 3x{y^2} + x{y^2}} \right) + 3{y^3} - 4\]

\[ = 2{x^3} + 3{y^3} - 4\].

Hướng dẫn giải

a) Ta có \(G = E + F\)\( = \left( {{x^7} - 4{x^3}{y^2} - 5xy + 7} \right) + \left( {{x^7} + 5{x^3}{y^2} - 3xy - 3} \right)\)

\( = {x^7} - 4{x^3}{y^2} - 5xy + 7 + {x^7} + 5{x^3}{y^2} - 3xy - 3\)

\( = \left( {{x^7} + {x^7}} \right) + \left( { - 4{x^3}{y^2} + 5{x^3}{y^2}} \right) + \left( { - 5xy - 3xy} \right) + \left( {7 - 3} \right)\)

\( = 2{x^7} + {x^3}{y^2} - 8xy + 4\).

Vậy \(G = 2{x^7} + {x^3}{y^2} - 8xy + 4.\)

b) Ta có \(E + H = F\)

Suy ra \(H = F - E\)\( = \left( {{x^7} + 5{x^3}{y^2} - 3xy - 3} \right) - \left( {{x^7} - 4{x^3}{y^2} - 5xy + 7} \right)\)

\( = {x^7} + 5{x^3}{y^2} - 3xy - 3 - {x^7} + 4{x^3}{y^2} + 5xy - 7\)

\( = \left( {{x^7} - {x^7}} \right) + \left( {5{x^3}{y^2} + 4{x^3}{y^2}} \right) + \left( { - 3xy + 5xy} \right) + \left( { - 3 - 7} \right)\)

\( = 9{x^3}{y^2} + 2xy - 10\).

Vậy \(H = 9{x^3}{y^2} + 2xy - 10.\)

a) \( - 3{x^2}{y^5}z:\left( {15x{y^3}} \right)\)

b) \(\left( { - 9{x^2}{y^3} + 6{x^3}{y^2} - 4x{y^2}} \right):3x{y^2}\)

\( =  - 9{x^2}{y^3}:3x{y^2} + 6{x^3}{y^2}:3x{y^2} - 4x{y^2}:3x{y^2}\)

c) \(\frac{1}{2}xy\left( {{x^5} - {y^3}} \right) - {x^2}y\left( {\frac{1}{4}{x^4} - {y^3}} \right)\)

\( = \frac{1}{2}xy \cdot {x^5} - \frac{1}{2}xy \cdot {y^3} - {x^2}y \cdot \frac{1}{4}{x^4} + {x^2}y \cdot {y^3}\)

\( = \frac{1}{2}{x^6}y - \frac{1}{2}x{y^4} - \frac{1}{4}{x^6}y + {x^2}{y^4}\)

\( = \left( {\frac{1}{2}{x^6}y - \frac{1}{4}{x^6}y} \right) - \frac{1}{2}x{y^4} + {x^2}{y^4}\)

d) \(\left( {x + 3y} \right)\left( {x - 2y} \right) - \left( {{x^4}y - 6{x^2}{y^3}} \right):{x^2}y\)

\( = {x^2} - 2xy + 3xy - 6{y^2} - \left( {{x^2} - 6{y^2}} \right)\)

\( = {x^2} + xy - 6{y^2} - {x^2} + 6{y^2}\)

\( = xy\).

Hướng dẫn giải

Ta có:

a) \(16{x^2} - 16x + 4 = 0\)

\(4{x^2} - 4x + 1 = 0\)

\({\left( {2x - 1} \right)^2} = 0\)

\(2x - 1 = 0\)

\(x = \frac{1}{2}\)

Vậy \(x = \frac{1}{2}\).

b) \[{\left( {x + 3} \right)^2} + \left( {5 - x} \right)\left( {5 + x} \right) = - 3\]

\[{x^2} + 6x + 9 + 25 - {x^2} = - 3\]

\[6x + 9 + 25 = - 3\]

\[6x = - 37\]

\(x = \frac{{ - 37}}{6}\)

Vậy \(x = \frac{{ - 37}}{6}\).

c) \[{x^3} + 9{x^2} + 27x + 19 = 0\]

\[{x^3} + 9{x^2} + 27x + 27 - 8 = 0\]

\[{\left( {x + 3} \right)^3} = 8\]

Suy ra \[x + 3 = 2\]

\(x =  - 1.\)

d) $25{\left(x+3\right)}^2+\left(1–5x\right)\left(1+5x\right)=8.$

\(25\left( {{x^2} + 6x + 9} \right) + \left[ {{1^2} - {{\left( {5x} \right)}^2}} \right] = 8\)

\[25{x^2} + 150x + 225 + 1 - 25{x^2} = 8\]

\[150x =  - 218\]

\(x =  - \frac{{109}}{{75}}.\)

Hướng dẫn giải

a) \(3{x^2} - \sqrt 3 x + \frac{1}{4}\)

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

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

b) \({x^2} - x - {y^2} + y\)

\( = \left( {{x^2} - {y^2}} \right) - \left( {x - y} \right)\)

\( = \left( {x - y} \right)\left( {x + y} \right) - \left( {x - y} \right)\)

\( = \left( {x - y} \right)\left( {x + y - 1} \right)\).

c) \[{x^4} + {x^3} + 2{x^2} + x + 1\]

\[ = \left( {{x^4} + 2{x^2} + 1} \right) + \left( {{x^3} + x} \right)\]

\[ = \left[ {{{\left( {{x^2}} \right)}^2} + 2{x^2} + 1} \right] + \left( {{x^3} + x} \right)\]

\[ = {\left( {{x^2} + 1} \right)^2} + x\left( {{x^2} + 1} \right)\]

d) \({x^3} + 2{x^2} + x - 16x{y^2}\)

\( = x\left( {{x^2} + 2x + 1 - 16{y^2}} \right)\)

\( = x\left[ {\left( {{x^2} + 2x + 1} \right) - {4^2}.{y^2}} \right]\)

\( = x\left[ {{{\left( {x + 1} \right)}^2} - {{\left( {4y} \right)}^2}} \right]\)

Lời giải:

Ta có \(VT = \left( {x - y} \right)\left( {{x^4} + {x^3}y + {x^2}{y^2} + x{y^3} + {y^4}} \right)\)

          \( = x\left( {{x^4} + {x^3}y + {x^2}{y^2} + x{y^3} + {y^4}} \right) - y\left( {{x^4} + {x^3}y + {x^2}{y^2} + x{y^3} + {y^4}} \right)\)

          \( = {x^5} + {x^4}y + {x^3}{y^2} + {x^2}{y^3} + x{y^4} - {x^4}y - {x^3}{y^2} - {x^2}{y^3} - x{y^4} - {y^5}\)

          \( = \left( {{x^5} - {y^5}} \right) + \left( {{x^4}y - {x^4}y} \right) + \left( {{x^3}{y^2} - {x^3}{y^2}} \right) + \left( {{x^2}{y^3} - {x^2}{y^3}} \right) + \left( {x{y^4} - x{y^4}} \right)\)

          \( = {x^5} - {y^5} = VP\).

Suy ra điều phải chứng minh.

Lời giải:

a) Ta có \[A = 4\left( {x - 2} \right)\left( {x + 1} \right) + {\left( {2x - 4} \right)^2} + {\left( {x + 1} \right)^2}\]

          \( = {\left( {2x - 4} \right)^2} + 2.2\left( {x - 2} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2}\)

          \( = {\left( {2x - 4} \right)^2} + 2.\left( {2x - 4} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2}\)

          \( = {\left[ {\left( {2x - 4} \right) + \left( {x + 1} \right)} \right]^2}\)

          \( = {\left( {2x - 4 + x + 1} \right)^2}\)

          \( = {\left( {3x - 3} \right)^2}\)

          \( = {\left[ {3\left( {x - 1} \right)} \right]^2}\)

          \( = 9{\left( {x - 1} \right)^2}\).

Do đó \(A = 9{\left( {x - 1} \right)^2}\).

Thay \[x = \frac{1}{2}\] vào \(A\) ta được \(A = 9{\left( {\frac{1}{2} - 1} \right)^2} = 9.{\left( { - \frac{1}{2}} \right)^2} = 9.\frac{1}{4} = \frac{9}{4}\).

Vậy \(A = \frac{9}{4}\) tại \[x = \frac{1}{2}\].

b) Ta có \[B = {x^9} - {x^7} - {x^6} - {x^5} + {x^4} + {x^3} + {x^2} - 1\]

          \( = \left( {{x^9} - {x^7}} \right) - \left( {{x^6} + {x^5}} \right) + \left( {{x^4} + {x^3}} \right) + \left( {{x^2} - 1} \right)\)

          \( = {x^7}\left( {{x^2} - 1} \right) - {x^5}\left( {x + 1} \right) + {x^3}\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)\)

          \( = {x^7}\left( {x - 1} \right)\left( {x + 1} \right) - {x^5}\left( {x + 1} \right) + {x^3}\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)\)

          \( = \left( {x + 1} \right)\left[ {{x^7}\left( {x - 1} \right) - {x^5} + {x^3} + \left( {x - 1} \right)} \right]\)

          \( = \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - \left( {{x^5} - {x^3}} \right)} \right]\)

          \( = \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - {x^3}\left( {{x^2} - 1} \right)} \right]\)

          \( = \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - {x^3}\left( {x - 1} \right)\left( {x + 1} \right)} \right]\)

          \( = \left( {x + 1} \right)\left( {x - 1} \right)\left[ {\left( {{x^7} + 1} \right) - {x^3}\left( {x + 1} \right)} \right]\)

          \( = \left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^7} + 1 - {x^4} - {x^3}} \right)\)

          \( = \left( {x + 1} \right)\left( {x - 1} \right)\left[ {\left( {{x^7} - {x^4}} \right) - \left( {{x^3} - 1} \right)} \right]\)

          \( = \left( {x + 1} \right)\left( {x - 1} \right)\left[ {{x^4}\left( {{x^3} - 1} \right) - \left( {{x^3} - 1} \right)} \right]\)

          \( = \left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^3} - 1} \right)\left( {{x^4} - 1} \right)\)

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

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

          \( = {\left( {x + 1} \right)^2}{\left( {x - 1} \right)^3}\left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)\).

Do đó \(B = {\left( {x + 1} \right)^2}{\left( {x - 1} \right)^3}\left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)\).

Thay \(x = 1\) vào \(B\) ta được \(B = {\left( {1 + 1} \right)^2}{\left( {1 - 1} \right)^3}\left( {{1^2} + 1} \right)\left( {{1^2} + 1 + 1} \right) = 0\).

Vậy \(B = 0\) tại \(x = 1\).