Ta có: \(x^2-2y^2=xy\)

\(\Leftrightarrow x^2-xy-2y^2=0\)

\(\Leftrightarrow x^2-2xy+xy-2y^2=0\)

\(\Leftrightarrow x\left(x-2y\right)+y\left(x-2y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Vì \(x+y\ne0\) nên x-2y=0

hay x=2y

Thay x=2y vào biểu thức \(A=\dfrac{x-y}{x+y}\), ta được: 

\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

Vậy: \(A=\dfrac{1}{3}\)

a)Ta có:\(x-y=2\Rightarrow\left(x-y\right)^2=4\Rightarrow\left(x^2+y^2\right)-2xy=4\Rightarrow4-2xy=4\Rightarrow2xy=0\Rightarrow xy=0\)

Khi đó ta có:\(x^5y=xy^5=xy\left(x^4-y^4\right)=0\)

x2 - 5x - 2xy + 5y + y2 + 4

= (x2 - 2xy + y2) - (5x - 5y) + 4

= (x2 - xy - xy + y2) - 5.(x - y) + 4

= (x - y)2 - 5.1 + 4

= 1 - 5 + 4

= 0

A=x2y−y+xy2−xx2y−y+xy2−x

A=(x2y−y)+(xy2−x)(x2y−y)+(xy2−x)

A=y(x2−1)+x(y2−1)y(x2−1)+x(y2−1)

A=y(x-1)(x+1)+x(y-1)(y+1)

thay x=-5 và y=2 ta có:

A=2(-5-1)(-5+1) - 5(2-1)(2+1)

A=2 . (-6) . (-4) - 5 . 3 

A=48 - 15

A=  33

\(x^2.y-y+x.y^2-x=\left(-5\right)^2.2-2+\left(-5\right).2^2-\left(-5\right)\)

\(=25.2-2-5.4+5=50-2-20+5=33\)

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

câu a đâu

\(A=x^2+y^2-xy^2-x^2y+2xy-5\)

\(=\left(x+y\right)^2-xy\left(y+x\right)-5\)

\(=2^2-2xy-5=-\left(2xy+1\right)\)

Trả lời:

\(A=x^2+y^2-x^2y-xy^2+2xy-5\)

\(A=\left(x^2+2xy+y^2\right)-xy.\left(x+y\right)-5\)

\(A=\left(x+y\right)^2-xy.\left(x+y\right)-5\)

\(A=2^2-xy.2-5\)

\(A=4-2xy-5\)

\(A=-1-2xy\)

\(A=-\left(1+2xy\right)\)

Học tốt 

1.a) xy + 2y - x² + 4

= y ( x + 2 ) - ( x² - 4 ) = y ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 )( y - x + 2 )

b) 2x² + y² + 3xy

= ( 2x² + 2xy ) + ( y² + xy )

= 2x ( x + y ) + y ( x + y )

= ( x + y ) ( 2x + y )

2.

x - y = 5 \(\Rightarrow\)( x - y )² = 25 \(\Rightarrow\)x² + y² = 25 + 2xy = 25 + 2.3 = 31

A = ( x + y )² = x² + y² + 2xy = 31 + 6 = 37