x 2 +y 2 xy ​ = 8 5 ​ ⇒x 2 +y 2 = 5 8xy ​ \Rightarrow P=\frac{\frac{8xy}{5}-2xy}{\frac{8xy}{5}+2xy}=\frac{8xy-10xy}{8xy+10xy}=\frac{-2}{18}=-\frac{1}{9}⇒P= 5 8xy ​ +2xy 5 8xy ​ −2xy ​ = 8xy+10xy 8xy−10xy ​ = 18 −2 ​ =− 9 1 ​

\(\frac{10}{3}\)

1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)

Đợi anh chút

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

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

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

a) Ta có: \(3x\left(2x-4\right)-\left(6x-1\right)\left(x+2\right)=25\)

\(\Rightarrow6x^2-12x-\left(6x^2+12x-x-2\right)=25\)

\(\Rightarrow6x^2-12x-6x^2-12x+x+2=25\)

\(\Rightarrow-23x+2=25\)

\(\Rightarrow-23x=25-2-23\)

\(\Rightarrow x=23:\left(-23\right)=-1\)

Vậy x = -1

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

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

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

\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

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

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

\(=\dfrac{x-y+z}{x-y-z}\)

Bài 1:

b) \(A+B=x^2+y^2+2x+3+2x^2+y^2+2x+1=3x^2+2y^2+4x+4\)

\(A-B=x^2+y^2+2x+3-2x^2-y^2-2x-1=-x^2+2\)

a) Ta có: \(A=x^2+y^2-2xy+2x+2xy+3\)

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

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

Ta có: \(B=2x^2+y^2-xy+2x+xy+1\)

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

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