ta có \(2x^2+2xy+2y^2+2x-2y+2=0\)

 <=>\(x^2+2xy+y^2+x^2+2x+1+y^2-2y+1=0\)

  <=>\(\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

<=>\(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

thay vào, ta có M=\(0^{30}+\left(-1+2\right)^{12}+\left(1-1\right)^{2017}=1\)

Vậy M=1 

^_^

Ta có : \(\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-2\end{array}\right.\)

Vậy \(x\in\left\{1;-2\right\}\)

Đây giống bài lớp 6 hơn

(x-1)(x+2)=0

=>x-1=0 hoặc x+2=0

=>x=1 hoặc x=-2

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

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

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

A = 2(x^2 - y^2).(x^4 + x^2y^2 + y^4) - 3x^4 - 3y^4 +1

A = 2x^4 + 2.x^2y^2 + 2y^4 - 3x^4 - 3y^4 +1

A = -x^4 + 2.x^2y^2 -y^4 +1

A = - (x^2 - y^2) +1

A = -1 + 1 =0

a/ \(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)

\(=x^7+x+\frac{1}{x}+\frac{1}{x^7}-\left(x+\frac{1}{x}\right)=x^7+\frac{1}{x^7}\)

b/ Ta có:

\(\left(x+\frac{1}{x}\right)^2=49\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=49-2=47\)

\(\left(x+\frac{1}{x}\right)^3=343\)

\(\Leftrightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=343\)

\(\Leftrightarrow x^3+\frac{1}{x^3}=343-3.7=322\)

\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=47.322=15134\)

\(\Leftrightarrow x^5+\frac{1}{x}+x+\frac{1}{x^5}=15134\)

\(\Leftrightarrow x^5+\frac{1}{x^5}=15134-7=15127\)

a)\(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)=x^7+x+\frac{1}{x}+\frac{1}{x^7}-x-\frac{1}{x}\)

=\(x^7+\frac{1}{x^7}\)

\(x+\frac{1}{x}=7\)

=>\(x\left(x+\frac{1}{x}\right)=7x\)

=>\(^{x^2-7x+1=0}\)

=>\(x=\frac{7+3\sqrt{5}}{2};x=\frac{7-3\sqrt{5}}{2}loại\)

=>\(x^5+\frac{1}{x^5}=15127\)

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

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

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

\(P=\frac{1}{x-4}\cdot x^2-2x-8\)

P\(P=\frac{x^2+2x-4x+8}{x-4}\)

\(P=\frac{x\left(x+2\right)-4\left(x+2\right)}{x-4}\)

\(P=\frac{\left(x-4\right)\left(x+2\right)}{x-4}\)

\(P=x+2\)

2 ,\(x^2-9x+20=0\)

\(\Rightarrow x^2-4x-5x+20=0\)

\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

\(\orbr{\begin{cases}x=5\Rightarrow\\x=4\Rightarrow\end{cases}}\orbr{\begin{cases}P=7\\P=6\end{cases}}\)