Đặt \(A=\sqrt{8}+3\)   

\(B=6+\sqrt{2}\)

\(A-B=\sqrt{8}+3-\left(6+\sqrt{2}\right)\)   

\(=2\sqrt{2}+3-6-\sqrt{2}\)   

\(=\sqrt{2}-3\)   

\(=\sqrt{2}-\sqrt{9}< 0\)

Vậy A < B      

đk: \(\hept{\begin{cases}-2\left(2x-3\right)\ge0\\\frac{5x+4}{2}\ge0\\\frac{-5x+2}{-5}\ge0\end{cases}}\Rightarrow\hept{\begin{cases}2x-3\le0\\5x+4\ge0\\2-5x\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge-\frac{4}{5}\\x\le\frac{5}{2}\end{cases}}\Rightarrow-\frac{4}{5}\le x\le\frac{3}{2}\)

Vậy \(-\frac{4}{5}\le x\le\frac{3}{2}\)

a.\(ĐKXĐ:\hept{\begin{cases}x^2-2x\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-2\right)\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-1\end{cases}}}\)

b.\(M=\left(\frac{1}{x^2-2x}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)

\(=\left(\frac{1}{x\left(x-2\right)}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)

\(=\left(\frac{1}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right)\div\frac{2x+1}{x\left(x+1\right)}\)

\(=\frac{2x+1}{x\left(x-2\right)}\div\frac{2x+1}{x\left(x+1\right)}\)

\(=\frac{2x+1}{x\left(x-2\right)}.\frac{x\left(x+1\right)}{2x+1}=\frac{x\left(2x+1\right)\left(x+1\right)}{x\left(x-2\right)\left(2x+1\right)}=\frac{x+1}{x-2}\)

c.Để \(M>1\)thì

 \(\frac{x+1}{x-2}>1\)

c, Ta có : \(M>1\Rightarrow\frac{x+1}{x-2}>1\Leftrightarrow\frac{x+1}{x-2}-1>0\)

\(\Leftrightarrow\frac{x+1-x+2}{x-2}>0\Leftrightarrow\frac{3}{x-2}>0\)

\(\Rightarrow x-2>0\Leftrightarrow x>2\)vì 3 > 0 

d, Để M nguyên khi \(x+1⋮x-2\Leftrightarrow x-2+3⋮x-2\)ĐK : \(x\ne2\)

\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

2x² + 4xy + 2y² - 8z² 

= 2(x² + 2xy + y² - 4z²)

= 2[(x + y)² - (2z)²]

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

phân đa thức thành nhân tử nha

Ta có:\(a^2-b=b^2-c\)

\(\Leftrightarrow a^2-b^2=b-c\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\)

\(\Leftrightarrow a+b=\frac{b-c}{a-b}\)

\(\Leftrightarrow a+b+1=\frac{b-c}{a-b}+1\)

\(\Leftrightarrow a+b+1=\frac{a-c}{a-b}\)

Cmtt ta có:

\(\hept{\begin{cases}b^2-c=c^2-a\Leftrightarrow b+c+1=\frac{b-a}{b-c}\\c^2-a=a^2-b\Leftrightarrow c+a+1=\frac{c-b}{c-a}\end{cases}}\)

\(\Rightarrow\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)=\frac{a-c}{a-b}.\frac{b-c}{b-a}.\frac{c-b}{c-a}=-1\)

Cre:mạng 

=x³+4³

=x³+64

=x³-27y³

=x³-4³

=x³-64