Question

tỉm x biết

\(a-\dfrac{1}{a+1}=\dfrac{2}{a^2-1}\) với a là hằng số

Answer 1

`ĐKXD:a\ne+-1`

`a-\frac{1}{a+1}=\frac{2}{a^2-1}`

`=>\frac{a^2+a-1}{a+1}-\frac{2}{(a-1)(a+1)}=0`

`=>\frac{1}{a+1}(a^2+a-1-\frac{2}{a-1}=0`

`=>\frac{a^3-2a+1-2}{a-1}=0`

`=>a^3-2a-1=0`

`=>a^3-a-a-1=0`

`=>a(a^2-1)-(a+1)=0`

`=>a(a-1)(a+1)-(a+1)=0`

`=>(a+1)(a^2-a-1)=0`

`=>(a+1)(a^2-a+1/4-5/4)=0`

`=>(a+1)[(a-1/2)^2-(\frac{\sqrt{5}}{2})^2]=0`

`=>(a+1)(a-\frac{1+sqrt{5}}{2})(a+\frac{-1+\sqrt{5}}{2})=0`

mà `a\ne+-1` 

`=>{(a=\frac{1+\sqrt{5}}{2}),(a=\frac{1-\sqrt{5}}{2}):}`