a) x² +x -y² + y = ( x² -y² ) +(x+y)

                         = (x-y)(x+y) +(x+y)

                         =(x+y)( x-y+1)

b) 3x² +3y² -6xy -12 = 3(x² +y² - 2xy) -12

                                 =3 [ (x-y)² -4]

                                 = 3( x-y-2)(x-y+2)

a) x² + x - y² + y

= (x² - y²) + (x + y)

= (x + y) (x - y) + (x + y)

= x + y

b) 3x² + 3y² - 6xy - 12

= 3 (x² + y² - 2xy - 4)

= 3 [(x² - 2xy + y²) - 4]

= 3 [(x - y)² - 2²]

= 3 (x - y + 2) (x - y - 2)

(sai thì thôi)

a) 3x² .(2x² - 3yz + x³)= 6x⁴ - 6x²yz +3x⁵

b)(24x⁵ - 12x⁴ + 6x² ).6x² = 144x⁷ - 72x⁶ +36x⁴

a) 3x² . (2x² - 3yz + x³)

= 3x² . 2x² + 3x² . (- 3yz) + 3x² . x³

= 6x⁴ + (-9x²yz) + 3x⁵

= 6x⁴ - 9x²yz + 3x⁵

a) Theo mình thì chỉ min thôi nhé!

\(A=\frac{8x^2-1}{4x^2+1}+1+11=\frac{12x^2}{4x^2+1}+11\ge11\)

b)Bạn rút gọn lại giùm mìn, lười quy đồng lắm:(

a, Ta có :\(x^{8n}+x^{4n}+1=x^{8n}+2x^{4n}+1-x^{4n}\)

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

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

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

\(=\left[\left(x^{2n}+1\right)-\left(x^n\right)^2\right]\left(x^{4n}-x^{2n}+1\right)\)

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

\(\Leftrightarrow x^{8n}+x^{4n}+1⋮x^{2n}+x^n+1\left(\forall x\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+3\right)=15\left(1\right)\)

Đặt \(x^2+x+2=t\)thay vào (1) ta được:

\(\left(t-1\right)\left(t+1\right)=15\)

\(\Leftrightarrow t^2-1=15\)

\(\Leftrightarrow t^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}t=4\\t=-4\end{cases}\left(2\right)}\)

Thay \(t=x^2+x+2\)vào (2) ta được: 

\(\orbr{\begin{cases}x^2+x+2=4\\x^2+x+2=-4\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-x+2x-2=0\\x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x-1\right)+2\left(x-1\right)=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0;\forall x\left(loai\right)\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy \(x\in\left\{1;-2\right\}\)