Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)

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

\(\Rightarrow\frac{x^2+2x-x+2}{\left(x-2\right)x}=\frac{2}{x\left(x-2\right)}\)

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

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

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

Vậy \(x\in\left\{0;-1\right\}\)

Sửa: đk: \(x\ne0\)

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

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

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

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

\(\Rightarrow x^2+2x-x=0\)

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

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

\(\Rightarrow\left[\begin{matrix}x=0\left(loại\right)\\x-1=0\end{matrix}\right.\Rightarrow x=1\)

Vậy x = 1

Ta có :

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

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

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

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

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

=> x(x+1)=0

\(\Rightarrow\left\{\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

x \(\ne0\) => x+1=0

=> x=(-1)

Tick mik nha!!

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

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

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

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

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

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

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

Bạn chú ý cách viết phương trình.

Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.

\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)

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

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

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

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

\(=16\)

Phương trình đã cho trở thành

\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)

\(\Leftrightarrow x=216-1=215\)

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

ĐKXĐ: \(x\ne\pm2\)

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

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

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

\(\Leftrightarrow x^2+4x+4+x^2-2x-x+2-2x^2-4=0\)

\(\Leftrightarrow x+2=0\)

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

Vậy: \(x=\varnothing\)

\(\Rightarrow S=\left\{\varnothing\right\}\)

Giải

\(\frac{x+2}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\\ĐKXĐ:x\ne2;-2\\ MTC:\left(x-2\right)\left(x+2\right)\\ \frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \left(x+2\right)\left(x-2\right)+\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\\ x^2-4+x^2-3x+2=2x^2+4\\ x^2-4+x^2-3x+2-2x^2=4\\-3x=6\\ x=-2 \)

quy đồng ,bỏ mẫu ,rút gọn =X² +X=0

             X=0 và X=-1

11111111111111111111111111111111111111111111111111111111111111111111111111111111

<=> \(\frac{\left(x+2\right)\cdot\left(x+2\right)}{x\cdot\left(x+2\right)}\)-\(\frac{x^2+5x+4}{x\left(x+2\right)}\)=\(\frac{x\left(x+2\right)}{\left(x+2\right)\cdot\left(x+2\right)}\)

=> x^2+4x+4-x^2-5x-4=x^2+2x

=> -x=x^2+2x

=> x^2+3x=0

=>x*(x+3)=0

mình nhầm bạn ơi cách làm vẫn như thế nhưng vế phải thì là\(\frac{x^2}{x\cdot\left(x+2\right)}\)

=> x=-1(ĐKXĐ:x khác 0)

\(\frac{1}{x^2+3}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{1}{2}\left(27-\frac{1}{x+9}\right)\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}=27-\frac{1}{x+9}\)

Mà 

\(\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}\)

\(=\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)

\(=\frac{1}{x}-\frac{1}{x+9}\)

\(\Rightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)