Question

giải các phương trình sau 

1, \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)

2, \(\dfrac{1}{x-6}-\dfrac{2}{6+x}=\dfrac{3x+6}{x^2-36}\)

Answer 1

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

Suy ra: \(x^2+4x+4+2x-4=x^2\)

\(\Leftrightarrow6x=0\)

hay \(x=0\left(nhận\right)\)

2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)

Suy ra: \(x+6-2x+12=3x+6\)

\(\Leftrightarrow-x-3x=6-18=-12\)

hay \(x=3\left(nhận\right)\)

Answer 2

Lời giải:
1. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)

\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)

\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)

2. ĐKXĐ: $x\neq \pm 6$

PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)

\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)

\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)

 

Answer 3

1) \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)

\(\Leftrightarrow\dfrac{x+2}{x-2}+\dfrac{2}{x+2}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2\left(x-2\right)-x^2}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{x^2+2x2+2^2+2x-4-x^2}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{x^2-x^2+4x+2x+4-4}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{6x}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow6x=0\)

\(\Rightarrow x=0\)

2) \(\dfrac{1}{x-6}-\dfrac{2}{6+x}=\dfrac{3x+6}{x^2-36}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{2}{x+6}-\dfrac{\left(3x+6\right)}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{1\left(x+6\right)-2\left(x-6\right)-\left(3x+6\right)}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{x+6-2x+12-3x-6}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{x-2x-3x+6-6+12}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{-4x+12}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow-4x+12=0\)

\(\Leftrightarrow-4x=12\)

\(\Rightarrow x=3\)

 

Answer 4

1. \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\) (*)

ĐKXĐ: \(x\ne\pm2\)

(*) \(\Leftrightarrow\dfrac{\left(x+2\right)^2+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)

\(\Leftrightarrow\dfrac{x^2+6x}{x^2-4}=\dfrac{x^2}{x^2-4}\) \(\Leftrightarrow x^2+6x=x^2\)

\(\Leftrightarrow6x=0\Leftrightarrow x=0\) ( thỏa mãn đk )

Vậy \(S=\left\{0\right\}\)

2. \(\dfrac{1}{x-6}-\dfrac{2}{6+x}=\dfrac{3x+6}{x^2-36}\) (**)

ĐKXĐ: \(x\ne\pm6\)

(**) \(\Leftrightarrow\dfrac{\left(x+6\right)-2\left(x-6\right)}{\left(x-6\right)\left(x+6\right)}=\dfrac{3x+6}{x^2-36}\)

\(\Leftrightarrow\dfrac{18-x}{x^2-36}=\dfrac{3x+6}{x^2-36}\) \(\Leftrightarrow18-x=3x+6\)

\(\Leftrightarrow12-4x=0\) \(\Leftrightarrow x=\dfrac{12}{4}=3\) (thỏa mãn đk )

Vậy \(S=\left\{3\right\}\)