\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-2=x^2-2x\)

\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

Cho mình sửa lại nhé:

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-2=x-2\)

\(\Leftrightarrow x^2+x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

ĐKXĐ:  \(x\ge1\)

\(\Rightarrow\left(\sqrt{x-1}+\sqrt{2x+1}\right)^2=1\Leftrightarrow x-1+2x+1+2\sqrt{\left(x-1\right)\left(2x+1\right)}=1\Leftrightarrow3x+2\sqrt{2x^2-x-1}=1\) \(\Leftrightarrow2\sqrt{2x^2-x-1}=1-3x\Rightarrow\left(2\sqrt{2x^2-x-1}\right)^2=\left(1-3x\right)^2\Leftrightarrow8x^2-4x-4=9x^2-6x+1\) \(\Leftrightarrow x^2-2x+5=0\Leftrightarrow\left(x-1\right)^2+4=0\Leftrightarrow\left(x-1\right)^2=-4\) vô lí vì VT\(\ge0\) mà VP<0 \(\Rightarrow\) ko có x Vậy...

\(3x^4+4x^3-3x^2-2x+1=0\)

\(\Leftrightarrow3x^4+x^3-x^2+3x^3+x^2-x-3x^2-x+1=0\)

\(\Leftrightarrow x^2\left(3x^2+x-1\right)+x\left(3x^2+x-1\right)-\left(3x^2+x-1\right)=0\)

\(\Leftrightarrow\left(x^2+x-1\right)\left(3x^2+x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x-1=0\left(1\right)\\3x^2+x-1=0\left(2\right)\end{cases}}\)

\(\Leftrightarrow x_{1,2}=\frac{-1\pm\sqrt{5}}{2}\left(tm\right)\)

\(x_{1,2}=\frac{-1\pm\sqrt{13}}{6}\left(tm\right)\)

`x^2+2x+3>2`

`<=>x^2+2x+1>0`

`<=>(x+1)^2>0`

`<=>x+1 ne 0`

`<=>x ne -1`

`(x+5)(3x^2+2)>0`

Vì `3x^2+2>=2>0`

`=>x+5>0<=>x>-5`

c) Ta có: \(21x-10x^2+9< 0\)

\(\Leftrightarrow10x^2-21x-9>0\)

\(\Leftrightarrow x^2-\dfrac{21}{10}x-\dfrac{9}{10}>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{21}{20}+\dfrac{441}{400}>\dfrac{801}{400}\)

\(\Leftrightarrow\left(x-\dfrac{21}{20}\right)^2>\dfrac{801}{400}\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{3\sqrt{89}+21}{20}\\x< \dfrac{-3\sqrt{89}+21}{20}\end{matrix}\right.\)

Câu 1: 

a) Ta có: 7x+21=0

\(\Leftrightarrow7x=-21\)

hay x=-3

Vậy: S={-3}

b) Ta có: 3x-2=2x-3

\(\Leftrightarrow3x-2-2x+3=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

c) Ta có: 5x-2x-24=0

\(\Leftrightarrow3x=24\)

hay x=8

Vậy: S={8}

Câu 2: 

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

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

Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)

b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)

c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+3-3\right)\left(x+3+3\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)

Vậy: S={0;-3;-6}

\(a)\)

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

\(\Leftrightarrow x-x^2+1=3x+1\)

\(\Leftrightarrow x^2-2x=0\)

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

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

\(b)\)

\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)

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

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)

^{\(a,\sqrt{2x-1}=2\)}

\(\Rightarrow2x-1=4\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=\frac{5}{2}\)

\(b,\sqrt{2x-1}=x+1\)

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

\(\Rightarrow2x-1=x^2+2x+1\)

\(\Rightarrow x^2+2x-2x=-1-1\)

\(\Rightarrow x^2=-2VN\)

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

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

\(4+4x^2+8x+x=0\)

\(4+4x^2+9x=0\)

=> vô nghiệm