(x+2)(x-2)+ 3x^2= (2x+ 1)^2+ 2x
mn giúp mik vs ạ mik đg cần gấp
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Lời giải:
$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$
$=\frac{(x+2)(x^2-2x+4)(x+1)(x+2)}{(x-1)^2(1-x)(x+1)}$
$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
\(\dfrac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2=\dfrac{1}{2}x^4+x^2y^2+\dfrac{1}{2}y^4-2x^2y^2\\ =\dfrac{1}{2}x^4-x^2y^2+\dfrac{1}{2}y^4=\dfrac{1}{2}\left(x^4-2x^2y^2+y^4\right)\\ =\dfrac{1}{2}\left(x^2-y^2\right)^2\)
\(2\left(x^2+y^2\right)^2-2x^2y^2=2\left(x^4+2x^2y^2+y^4\right)-2x^2y^2\\ =2x^4+4x^2y^2+2y^4-2x^2y^2=2x^4+2x^2y^2+2y^4\\ =2\left(x^4+x^2y^2+y^4\right)\)
h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
`(x-2)(2x-1)=0`
\(=>\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
<=> x2 -4+3x2= 4x2+4x+1+2x
<=> 4x^2 - 4= 4x^2 +6x +1
<=> - 4=6x +1
<=> 6x= -5
<=> x= \(-\frac{5}{6}\)