đkxđ : x khác 1 ; x khác -1
ta có pttđ:
\(\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{4x}{x^2-1}-\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
<=>
\(\left(x+1\right)^2-4x-\left(x-1\right)^2=0\)
<=>
\(x^2+2x+1-4x-\left(x^2-2x+1\right)=0\)
<=>
\(x^2+2x+1-4x-x^2+2x-1=0\)
<=>
\(0=0\)
=> x thuộc R ngoài 1 và -1
\(\dfrac{X+1}{X-1}-\dfrac{4x}{x^2-1}=\dfrac{x-1}{x+1}\left(Đkx\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right).\left(x+1\right)}-\dfrac{4x}{\left(x-1\right).\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right).\left(x+1\right)}\)\(\Rightarrow x^2+2x+1-4x=x^2-2x+1\)
\(\Leftrightarrow x^2+2x-4x-x^2+2x=1-1\)
\(\Leftrightarrow0x=0\) (luôn đúng)
vậy phương trình có vô số nghiệm trừ x \(\ne1;x\ne-1\)
ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x+1}{x-1}-\dfrac{4x}{x^2-1}=\dfrac{x-1}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4x}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x}{x^2-1}\)
\(\Leftrightarrow\dfrac{x^2+2x+1-x^2+2x-1}{x^2-1}=\dfrac{4x}{x^2-1}\)
\(\Leftrightarrow4x=4x\)
Vậy với \(\forall x\ne\pm1\) thì phương trình thỏa mãn
