\(3,x=\dfrac{1}{2},y=-1\)

\(\Rightarrow C=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+1\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-1\right)-1\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

\(\Rightarrow C=\dfrac{1}{2}\left(\dfrac{1}{4}+1\right)-\dfrac{1}{4}\left(-\dfrac{1}{2}\right)-\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)

\(\Rightarrow C=\dfrac{1}{2}.\dfrac{5}{4}+\dfrac{1}{8}-\left(-\dfrac{1}{4}\right)\)

\(\Rightarrow C=\dfrac{5}{8}+\dfrac{1}{8}+\dfrac{1}{4}\)

\(\Rightarrow C=1\)

\(4,x=\dfrac{1}{2},y=-100\)

\(\Rightarrow D=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+100\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-100\right)-100\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

\(\Rightarrow D=\dfrac{1}{2}\left(\dfrac{1}{4}+100\right)-\dfrac{1}{4}\left(-\dfrac{199}{2}\right)-100\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)

\(\Rightarrow D=\dfrac{1}{2}.\dfrac{401}{4}+\dfrac{199}{8}-100.\left(-\dfrac{1}{4}\right)\)

\(\Rightarrow D=\dfrac{401}{8}+\dfrac{199}{8}+25\)

\(\Rightarrow D=100\)

3: C=x^3-xy-x^3-x^2y+x^2y-xy

=-2xy=-2*1/2*(-1)=1

4: D=x^3-xy-x^3-x^2y+x^2y-xy

=-2xy

=-2*1/2*(-100)=100

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

\(= x^2+2xy-y^2\) (a)

Thay \(x= - 8\) và \(y=7\) vào (a),ta có :

\(\left(-8\right)^2+2.\left(-8\right).7-7^2=64-112-49\)

\(= -97\)

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

\(= x^3-xy+x^2y-xy-x^3-xy^2\)

\(=-2xy\) (b)

Thay \(x=\dfrac{1}{2}\) và \(y=-100\) vào (b),ta có :

\(-2.\dfrac{1}{2}.\left(-100\right)=100\)

HỌC TỐT NHA !!!

a) x(x+y) + y(x-y)

= x.x + x.y +y.x - y.y

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

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

Thay x = -8 và y=7 vào biểu thức ta có :

\(\left(-8\right)^2+2.\left(-8\right).7+7^2\)

\(64+\left(-112\right)+49=1\)

Vậy ...

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

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

\(=\left(x^3-x^3\right)-\left(xy+xy\right)+\left(xy^2-xy^2\right)\)

\(=0-2xy+0=-2xy\)

Thay x = \(\frac{1}{2}\) và y = -100 ta có :

\(-2.\left(\frac{1}{2}\right).\left(-100\right)=100\)

Vậy ...

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

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

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

\(=\dfrac{-1}{x-2y}\)

Thay $x=-1,76$ và $y=\dfrac{3}{25}$ vào $P=\dfrac{-1}{x-2y}$, ta được:

$P=\dfrac{-1}{-1,76-2.(\dfrac{3}{25})}=\dfrac{1}{2}$.

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y\right)^2+\dfrac{1}{\left(x+y\right)^2}+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y+\dfrac{1}{x+y}\right)^2+\left(x-y+\dfrac{1}{x-y}\right)^2=\dfrac{136}{9}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=u\\x-y+\dfrac{1}{x-y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{16}{3}\\u^2+v^2=\dfrac{136}{9}\end{matrix}\right.\)

Hệ cơ bản, chắc bạn tự giải quyết phần còn lại được

a: \(=x^2-xy+xy+y^2=x^2+y^2=100\)

b \(=x^3-xy-x^3-x^2y+x^2y-xy=-2xy=-2\cdot\dfrac{1}{2}\cdot\left(-100\right)=-1\cdot\left(-100\right)=100\)

a)` x(x - y) + y(x + y) `

`=x^2-xy+xy+y^2`

`=x^2+y^2`(1)

thay x= -6 ; y= 8 vào 1 ta đc

\(\left(-6\right)^2+8^2=36+64=100\)

b)`) x(x^2 - y) - x^2 (x + y) + y (x^2 - x) `

`=x^3-xy-x^3-xy+yx^2-xy`

`=\(-3xy+yx^2\)(2)

thay `x= 1/2 và y = -100` ta đc

\(-\dfrac{3.1}{2}.\left(-100\right)+\dfrac{\left(-100\right).1}{2}=150-50=100\)

=100

a) \(=x^2-xy+xy+y^2=x^2+y^2\)tự thay rồi tính nha

b) \(=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\) tự thay vào nha

a) x(x – y) + y (x + y) = x² – xy +yx + y²= x²+ y²

với x = -6, y = 8 biểu thức có giá trị là (-6)² + 8² = 36 + 64 = 100

b) x(x² – y) – x² (x + y) + y (x²– x) = x³ – xy – x³ – x²y + yx² – yx= (2x-2y) – (x² -2xy +y²) =2(x-y) – (x-y)²

Với x =1/2, y = -100 biểu thức có giá trị là -2 . 1/2. (-100) = 100.

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

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

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

Tại \(x=-6;y=8\)thì biểu thức có giá trị là:

\(\left(-6\right)^2+8^2=100\)

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

\(=x^3-xy-x^3-x^2y+x^2y-xy\)

\(=-2xy\)

Tại \(x=\frac{1}{2};y=100\)

\(-2.\frac{1}{2}.100=-100\)