x² - 4x + 4 = ( x +1)² - 8x

x² - 4x + 4 = x² + 2x + 1 - 8x

x² - 4x + 4 - x² - 2x - 1 + 8x = 0

2x + 3 = 0

2x = -3

x = -3/2

\(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\\ =>-5\left(4x-1\right)\left(x-2\right)-2\left(4x-1\right)^2=0\\ =>\left(4x-1\right)\left[-5\left(x-2\right)-2\left(4x-1\right)\right]=0\\ =>\left(4x-1\right)\left(-5x+10-8x+2\right)=0\\ =>\left(4x-1\right)\left(-13x+12\right)=0\\ =>\left[{}\begin{matrix}4x-1=0\\-13x+12=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)

\(2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\\ 2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\\ \left(3x+1\right)\left(6x+2-x+2\right)=0\\ \left(3x+1\right)\left(5x+4\right)=0\\ \left[{}\begin{matrix}3x+1=0\\5x+4=0\end{matrix}\right.\\ \left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

2x - 4y = 0

<=> 2x = 4y 

<=> x/y=2

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

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

\(=\dfrac{-x^2+y^2+x^2-y^2}{xy\left(x-y\right)}=0\)

b: \(=\dfrac{x^3-1}{x-1}+\dfrac{-x^2+1}{x+1}\)

\(=x^2+x+1-x+1=x^2+2>0\)

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

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

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

\(\left|x+1\right|=3\\ \left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=2\left(loai\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

với x=-4 thì

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

\(=>P=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{-2x-4}{x^2-4}\)`(x ne +-2)`

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

\(P=\dfrac{x^2-2x+2x+4-2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(P=\dfrac{x}{x+2}\)

`|x+1| =3`

`=>[(x+1=3),(x+1=-3):}`

`=> [(x=3-1=2(ktm) ),(x=-3-1=-4(t/m)):}`

Thay `x=-4` vào `P` ta đc

`P= (-4)/(-4+2) = 2`

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

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

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

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

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

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

\(=\left(x-y\right)\left(x-y\right)\left(x-y-1\right)\)

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

\(=8^2\left(8-1\right)\\ =64\cdot7=448\)

sai đề ko bn?

đề có sai ko bn?