Bài 1 :

Tự phân tích vế trái và điền vào vế phải 

Bài 2 :

a) \(3x^3-6x^2+3x\)

\(=3x\left(x^2-2x+1\right)\)

\(=3x\left(x-1\right)^2\)

b) \(2xy+z+2x+yz\)

\(=\left(2xy+2x\right)+\left(z+yz\right)\)

\(=2x\left(y+1\right)+z\left(y+1\right)\)

\(=\left(y+1\right)\left(2x+z\right)\)

c) \(x^4-y^4\)

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

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

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

d) \(3x^2-4x-7\)

\(=3x^2+3x-7x-7\)

\(=3x\left(x+1\right)-7\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-7\right)\)

Gọi biểu thức trên là T. Ta có:

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

\(=x^2-6x+9+x^2-4x+4\)

\(=2x^2-10x+13\)

Ta có: \(T=2x^2-10x+13\)

\(=2x^2-10x+\frac{25}{2}+\frac{1}{2}\)

\(=2\left(x^2-5x+\frac{25}{4}\right)+\frac{1}{2}\)

\(=2\left(x-\frac{5}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\) (do \(2\left(x-\frac{5}{2}\right)^2\ge0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=0\Leftrightarrow x=\frac{5}{2}\)

Vậy \(T_{min}=\frac{1}{2}\Leftrightarrow x=\frac{5}{2}\)

\(\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)

\(\left(x^2+4x-5\right)\)\(\left(x^2+4x+5\right)\)

=\(\left(x^2+4x\right)^2-25\)

Min=-25 khi \(^{x^2+4x=0}\)

\(\Rightarrow x=0\) hoặc x=-4

Gọi biểu thức trên là A.Ta có:

\(A=3x^2-5x+7\)

\(=3x^2-5x+\frac{25}{12}+\frac{59}{12}\)

\(=3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)+\frac{59}{12}\)

\(=3\left(x-\frac{5}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}\) (do \(3\left(x-\frac{5}{6}\right)^2\ge0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{5}{6}\right)^2=0\Leftrightarrow x=\frac{5}{6}\)

Vậy \(A_{min}=\frac{59}{12}\Leftrightarrow x=\frac{5}{6}\)

a) \(x^2\left(x-3\right)+12-4x=0\)

\(x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\left(x-3\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x\in\left\{\pm2\right\}\end{cases}}\)

b) \(x\left(2x-7\right)-3\left(7-2x\right)=0\)

\(x\left(2x-7\right)+3\left(2x-7\right)=0\)

\(\left(2x-7\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-7=0\\x+3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-3\end{cases}}\)

c) \(\left(2x-1\right)^2-25=0\)

\(\left(2x-1\right)^2-5^2=0\)

\(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\left(2x-6\right)\left(2x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

d) \(\left(3x-5\right)^2-\left(2x-3\right)^2=0\)

\(\left(3x-5-2x+3\right)\left(3x-5+2x-3\right)=0\)

\(\left(x-2\right)\left(5x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x-8=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{5}\end{cases}}\)

\(P=\frac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{\left(x^2\right)^2-1}\)

\(=\frac{x^8\left(x^2-1\right)+x^4\left(x^2-1\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

\(=\frac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{x^8+x^4+1}{x^2+1}\)