bài 4: Ta có \(x^2-2y^2=xy\Rightarrow x^2-y^2=xy+y^2\Rightarrow\left(x-y\right)\left(x+y\right)=y\left(x+y\right)\)

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

thay x=2y vào A ta đc :

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

Bài 1:

Ta có: \(x+y+z=0\Rightarrow z=-x-y\Rightarrow z^2=(-x-y)^2\)

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

Hoàn toàn tương tự:

\(y^2+z^2-x^2=-2yz; z^2+x^2-y^2=-2xz\)

Do đó:

\(P=\frac{(x^2+y^2-z^2)(y^2+z^2-x^2)(z^2+x^2-y^2)}{16xyz}=\frac{(-2xy)(-2yz)(-2xz)}{16xyz}=\frac{-xyz}{2}\)

\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)

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

\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)

\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)

a: \(B=\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)\)

\(=x^2y+\dfrac{1}{4}x^2+y^2+\dfrac{1}{4}y+x^2y^2+\dfrac{3}{4}y+\dfrac{1}{4}\)

\(=x^2y+x^2y^2+y^2+y+\dfrac{1}{4}x^2+\dfrac{1}{4}\)

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

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

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

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

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

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

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

=>\(A=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)

b: \(=\dfrac{y^2-y+1+2y-\dfrac{3}{4}}{y^2-y+1}=1+\dfrac{2y-\dfrac{3}{4}}{y^2-y+1}>=1\)

Dấu = xảy ra khi y=3/8

a: \(P=\left(\dfrac{2y^2+1}{\left(y+1\right)\left(y^2-y+1\right)}-\dfrac{1}{y+1}\right):\dfrac{y^2-y+1-y^2+2y+1}{y^2-y+1}\)

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

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

b: |2y+5|=3

=>2y+5=3 hoặc 2y+5=-3

=>2y=-2 hoặc 2y=-8

=>y=-1(loại) hoặc y=-4(nhận)

Thay y=-4 vào P,ta được:

\(P=\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)

c: Để P chia hết cho 4 thì P=4k

=>y=4k(y+2)

a) ĐKXĐ: x \(\ne\) -2 và x \(\ne\) 2

S=\(\left\{x|-2< x< 2\right\}\)