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

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

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

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

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

\(\Rightarrow C=1\)

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

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

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

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

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

\(\Rightarrow D=100\)

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

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

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

\(P=\left(\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\right):\frac{2x^2+y+2}{x+1}\)

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

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

Đặt \(A=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\)

      \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

A = \(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-y^2-xy-y^2}=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)

=>A=\(\frac{x^2-y^2+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

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

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

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

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

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

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

\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)

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

2: \(\left(x^2-y^2\right)\cdot C=-8\)

=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)

=>\(\left(x-y\right)^3=-8\)

=>x-y=-2

=>x=y-2

\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)

\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)

\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)

\(=\left(y-1\right)\left(-4y+4\right)+4xy\)

\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)

\(=-4y^2+8y-4+4y^2-8y\)
=-4

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

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

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

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

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

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

2x² + 2y² + 3xy - x + y + 1 = 0

2x² + 2y² + 4xy - xy - x + y + 1 = 0

(2x² + 2y² + 4xy) + (-xy - x) + (y + 1) = 0

2(x + y)² - x(y + 1) + (y + 1) = 0

2(x + y)² + (y + 1)(1 - x) = 0

Do (x + y)² \(\ge0\)

\(\Rightarrow\) 2(x + y)² \(\ge0\)

\(\Rightarrow\) 2(x + y)² + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0

\(\Rightarrow y+1=0;1-x=0\)

*) y + 1 = 0

y = -1

*) 1 - x = 0

x = 1

Với x = 1; y = -1, ta có:

B = [1 + (-1)]²⁰¹⁸ + (1 - 2)²⁰¹⁸ + (-1 - 1)²⁰¹⁸

= 1 + 2²⁰¹⁸