(3x-1)² -7(x²+2)= 9x^2-6x+1-7x^2-28x-28= 2x^2-34x-27

xin !

Bài 1 . 

\(a,3x^2-6x=3x\left(x-2\right)\)

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

Bài 2

\(a,x\left(x-1\right)-x^2+2x=5\)

\(x^2-x-x^2+2x=5\)

\(x=5\)

\(b,4x^3-36x=0\)

\(4x\left(x^2-9\right)=0\)

\(4x\left(x-3\right)\left(x+3\right)=0\)

TH1 : x = 0

TH2 : x - 3 = 0  => x = 3

TH3 : x + 3 = 0 => x = -

Các câu mình chưa làm thì bạn t lm nh 

Bài 1. Phân tích các đa thức sau thành nhân tử:

\(3x^2-6x\)

\(=3x.\left(x-2\right)\)

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

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

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

\(9x^3-9x^2y-4x+4y\)

\(=9x^2.(x-y)-4.\left(x-y\right)\)

\(=\left(x-y\right).\left(9x^2-4\right)\)

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

\(x^3-2x^2-8x\)

\(=x.\left(x^2-2x-8\right)\)

\(=x.\left(x^2+2x-4x-8\right)\)

\(=x.[x.\left(x+2\right)-4.\left(x+2\right)]\)

\(=x.\left(x+2\right).\left(x-4\right)\)

TL:

Hình đâu bạn

HT

hình vẽ đấu bạn ơi

a) \(\left(x+1\right)^2-4\left(x+2\right)^2=0\Leftrightarrow\left(x+1\right)^2-\left[2\left(x+2\right)\right]^2=0\)

\(\Leftrightarrow\left[x+1-2\left(x+2\right)\right]\left[x+1+2\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x+1-2x-4\right)\left(x+1+2x+4\right)=0\)

\(\Leftrightarrow\left(-x-3\right)\left(3x+5\right)=0\Leftrightarrow\orbr{\begin{cases}-x-3=0\\3x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=3\\3x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-\frac{5}{3}\end{cases}}}\).    Vậy .......

b) \(\left(x+2\right)^2-x^2-4=0\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\).   Vậy x = -2

c) \(x+\sqrt{x}-12=0\Leftrightarrow x+4\sqrt{x}-3\sqrt{x}-12=0\) ( ĐKXĐ : \(x\ge0\) ) 

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+4\right)-3\left(\sqrt{x}+4\right)=0\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)=0\)

\(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+4\ge4>0\Rightarrow\sqrt{x}+4\ne0\)

\(\Rightarrow\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(TMĐK\right)\).

Vậy \(x=9\)

Ban giup Tôi giai Bài 1+1=?

TL:

=2

-HT-

 Đáp án là :1 + 1 = 2