Giải pt: `8x^2-1 =2x\sqrt{2x+3}`

`8x^2 -1= 2x` $\sqrt[]{2x+3}$ 

`⇔ 9x^2 -1+2x+3= x ^2 +2x `$\sqrt[]{2x+3}$  `+2x+3`

`⇔ 9x^2 +2x+2=` `( x +x` $\sqrt[]{2x+3}$  `)^2`

`⇔  2(x+1)=` `( x +x` $\sqrt[]{2x+3}$  `)^2` `-(3x)^2` 

`⇔  2(x+1)=` `-x^2(2- `$\sqrt[]{2x+3}$ `)(` $\sqrt[]{2x+3}$ `+4)`

Đặt `t=` $\sqrt[]{2x+3}$ 

`⇒ t^2-1 = [ 2(t^3-3)+t][t-t^2+3]`

`⇒ t^2-1 = -2t^4+2t^3+6t^2-t^3+t^2+3t+6t^2-6t-18`

`⇒ -2t^4+t^3+12t^2-3t-17 = 0`