Question

giúp mình với mai mink nộp bài

giải phương trình sau

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

Answer 1

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 \(x\in\left\{0;-1\right\}\)

Answer 2

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

Answer 3

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!!

Answer 4

<=>\(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\) (đkxđ :x\(\ne\)2)

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

<=>x²+2x-x+2=2

<=>x²+2x-x=2-2

<=>x²-x=0

<=>x(x-1)=0

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

vậy phương trình có tập nghiệp S={0,1}