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A = x ( x + y ) - y ( x + y )

A = ( x + y ) ( x - y )

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

\(A=x\left(x+y\right)-y\left(x+y\right)\)

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

\(=x^2-y^2\)

Thay \(x=-\dfrac{1}{2}\) và \(y=-2\) vào biểu thức \(A\), ta có:

\(A=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)

\(=\dfrac{1}{4}-4\)

\(=-\dfrac{15}{4}\)

sửa đề : bạn check lại đề xem nhé

 \(A=\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(y-x\right)\)

\(=\left(x+y\right)^2-2\left(x+y\right)\left(y-x\right)+\left(y-x\right)^2\)

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

Thay x = -1 ; y = -2 ta được : \(4.1=4\)

\(A=\left(x+y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(y-x\right)\)

\(=2\left(x+y\right)^2-2\left(x+y\right)\left(y-x\right)=2\left(x+y\right)\left[\left(x+y\right)-\left(y-x\right)\right]\)

\(=2\left(x+y\right)\left(x+y-y+x\right)=2.2x\left(x+y\right)=4x\left(x+y\right)\)

Thay x = -1 ; y = -2 ta được : \(-4.\left(-3\right)=12\)

26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)

Thay x = 15 vào bt A ta có

A = 9 . 15 = 135

b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)

Thay x = -1/5 ; y = - 1/2 vào bt B ta có

\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(=9x^2y^2-xy^3-8x^3\)

Thay x = 1/2 ; y = 2 vào bt C ta có

\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)

d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)

\(=12x^2+12x-3\)

\(\left|x\right|=2\Rightarrow x=\pm2\)

Thay x = 2 vào bt D có

\(D=12.4+12.2-3=69\)

Thay x = - 2 vào bt D ta có

\(D=12.4-12.2-3=21\)

Bài 1 :

a) \(M=\dfrac{1}{2}x^2y.\left(-4\right)y\)

\(\Rightarrow M=-2x^2y^2\)

Khi \(x=\sqrt[]{2};y=\sqrt[]{3}\)

\(\Rightarrow M=-2.\left(\sqrt[]{2}\right)^2.\left(\sqrt[]{3}\right)^2\)

\(\Rightarrow M=-2.2.3=-12\)

b) \(N=xy.\sqrt[]{5x^2}\)

\(\Rightarrow N=xy.\left|x\right|\sqrt[]{5}\)

\(\Rightarrow\left[{}\begin{matrix}N=xy.x\sqrt[]{5}\left(x\ge0\right)\\N=xy.\left(-x\right)\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}N=x^2y\sqrt[]{5}\left(x\ge0\right)\\N=-x^2y\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)

Khi \(x=-2< 0;y=\sqrt[]{5}\)

\(\Rightarrow N=-x^2y\sqrt[]{5}=-\left(-2\right)^2.\sqrt[]{5}.\sqrt[]{5}=-4.5=-20\)

2:

Tổng của 4 đơn thức là;

\(A=11x^2y^3+\dfrac{10}{7}x^2y^3-\dfrac{3}{7}x^2y^3-12x^2y^3=0\)

=>Khi x=-6 và y=15 thì A=0

\(Q=x\left(x^2+y\right)-x^2\left(x+y\right)+y\left(x^4+x\right)\)

=> \(Q=\left(x^3+xy\right)-\left(x^3+x^2y\right)+\left(x^4y+xy\right)\)

=> \(Q=\left(x^3-x^3\right)+\left(xy+xy\right)+\left(x^4y-x^2y\right)\)

=> \(Q=x^4y-x^2y+2xy\)

=> \(Q=\frac{2^4.1}{2}-\frac{2^2.1}{2}+\frac{2.2.1}{2}\)

=> \(Q=2^3-2+2=2^3=8\)

Vậy \(Q=8\)