a) x+5x²=0

<=> x(1+5x)=0

<=>\(\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)

b) x+1=(x+1)²

<=> x+1-(x+1)²=0

<=> (x+1)(1-x-1)=0

<=> -x(x+1)=0

<=> \(\hept{\begin{cases}-x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)

c) x³+x=0

<=> x(x²+1)=0

<=> x=0 (vì x²+1>0)

Ta có : (x - 2)(x² + 4) = x² - 2x 

<=> (x - 2)(x² + 4) = x(x - 2) 

<=> (x - 2)(x² - x + 4) = 0

<=> x - 2 = 0 (vì \(x^2-x+4=\left(x-\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\forall x\))

<=> x = 2

Vậy x = 2 

Ta có : (x + 1)(2x - 3) \(\le\)0

 Xét 2 trường hợp : 

TH1 : \(\hept{\begin{cases}x+1\le0\\2x-3\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\le-1\\x\ge1,5\end{cases}}\left(\text{loại}\right)\)

TH2 : \(\hept{\begin{cases}x+1\ge0\\2x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le1,5\end{cases}}\Leftrightarrow-1\le x\le1,5\)

Vậy \(-1\le x\le1,5\)

 y-x^2y-2xy^2-y^3
=y(1-x^2+2xy-y^2)
=y[1^2-(x-y)^2]
=y(1-x+y)(1+x-y)

x² - 2xy + y² - xz + yz

= ( x - y )² - z( x - y )

= ( x - y )( x- y - z )

a, Với \(x\ne\pm3\)

\(P=\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{3}{x+3}\)

\(=\frac{6x+5x\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x+5x^2+15x+3x-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{5x^2+24x-9}{\left(x+3\right)\left(x-3\right)}\)hmm bạn xem lại biểu thức trên nhé 

a)(x-3)²+x(2-x)=7

x²-2*3x+3²+2x-x²=7

x²-6x+2x-x²=7+3²

x²-x²-6x+2x=16

-4x=16

-x= -4

b)...

12x-4x²+4x²+2x-2x-1=5

12x-4x²+4x²+2x-2x=5+1

-4x²+4x²+12x+2x-2x=6

12x=6

x=\(\frac{6}{12}\)

c)2x²-3x-9=0

x(2x-3)-9=0

[_2x-3=0^x-9=0

[^x=9_{x=\(\frac{3}{2}\)}

_{câu b mình ko ghi đè nên thông cảm}

_{nhớ k cho mình nha ^ ^}

2) x³ - x² - 4x + 4

= x² (x - 1) - 4(x - 1)

= (x² - 4)(x - 1)

= (x-2)(x+2)(x-1)

HT