\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\)              \(\left(x\ne1;x\ne-2\right)\)

\(\Rightarrow\frac{2\left(x-1\right)+2}{x-1}=\frac{\left(x+2\right)+2x}{x+2}\)\(\Rightarrow2x^2+4x=3x^2+2x-3x+2\)

\(\Rightarrow\frac{2x-2+2}{x-1}=\frac{x+2+2x}{x+2}\)

\(\Rightarrow\frac{2x}{x-1}=\frac{3x+2}{x+2}\)

\(\Rightarrow2x\left(x+2\right)=\left(x-1\right)\left(3x+2\right)\)

\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)

\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)

\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ne-2\end{cases}}\)

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

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

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

=> \(2x\left(x+2\right)=x+2\left(x+1\right)\left(x-1\right)\)

=> \(2x^2+4x=x+2\left(x^2-1\right)\)

=> \(2x^2+4x=x+2x^2-2\)

=> \(2x^2+4x-x-2x^2+2=0\)

=> \(3x+2=0\)

=> \(3x=-2\)

=> \(x=-\frac{2}{3}\)

We have  \(3x^2+8x^3+x^4+9-8x^3-3x^2\)

\(=\left(3x^2-3x^2\right)+\left(8x^3-8x^3\right)+\left(x^4-9\right)\)

\(=x^4-9\)

If my answer is right, I hope you k for me =))    =.='

key: \(3x^2+8x^3+x^4+9+\left(-2x\right)^3-3x^2\)

\(=3x^2+8x^3+x^4+9-2x^3-3x^2\)

\(=\left(3x^2-3x^2\right)+\left(8x^3-2x^3\right)+x^4+9\)

\(=4x^3+x^4+9\)

Bạn Hương sai rồi : đề bài người ta cho là \(\left(-2x\right)^3=-8x^3\)

chứ ko phải là \(\left(-2x^3\right)\)

Với lại \(\left(8x^3-2x^3\right)=6x^3\)chứ đâu phải là \(4x^3\)

=>(x-2)(x+2)+3(x-2)=0

=>(x-2)(x+5)=0

=>x=2 hoặc x=-5

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\) 

theo đề ta có\(x^2-4+3\left(x-2\right)=0\)

\(x^2-4+3x-6=0\)

\(\Leftrightarrow x^2+3x=6+4=10\)

\(x\left(x+3\right)=10\)

\(\left\{{}\begin{matrix}x=10\\x+3=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\x=7\end{matrix}\right.\)

vậy x \(\in\left\{10,7\right\}\)

x + 1 = 2009 => x = 2008

=> (x+1).(x+1)=2009.2009

(x+1)²=4036081

(x+1)²=2009²

^X+1=2009

^x=2009-1

^x=2008

^hoặc

^x+1=-2009

^x=-2009-1

^x=-2010

^3 la j

\(x\in\left(\infty;-\infty\right)\)

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

\(-\left(x-1\right)^3=2^5.3\)

\(1-2\sqrt[3]{12}\)

Sau đó bạn tự\(\Rightarrow\)X nha

5^x+3(5-1*3)=5^11*2

5^x+3*2=5^11*2

5^x+3=5^11

x+3=11

x=8

=>-2x+7-3x^2-x=0

=>-3x^2-3x+7=0

=>\(x=\dfrac{-3\pm\sqrt{93}}{6}\)

\(c.-\left(2x-7\right)-x\left(3x+1\right)=0\) 

\(\Leftrightarrow-2x+7-3x^2-x=0\) 

\(\Leftrightarrow-3x^2-3x+7=0\) 

Vậy pt này vô n₀ 

Sai đề bạn ơi

Sai chỗ \(\frac{1}{x^2+6x+18}\)

Bạn sửa lại rồi mk giải cho..