2 $\sqrt[2]{2x-1}$+ $\sqrt[2]{x+3}$ -$\sqrt[2]{5x+11}$ =0 03/11/2021 Bởi Faith 2 $\sqrt[2]{2x-1}$+ $\sqrt[2]{x+3}$ -$\sqrt[2]{5x+11}$ =0
`ĐKXĐ: x\ge 1/2` `2\sqrt(2x-1)+\sqrt(x+3)-\sqrt(5x+11)=0` `⇔2\sqrt(2x-1)+\sqrt(x+3)=\sqrt(5x+11)` `⇔4(2x-1)+x+3+4\sqrt[(2x-1)(x+3)]=5x+11` `⇔4\sqrt[(2x-1)(x+3)]=12-4` `⇔\sqrt[(2x-1)(x+3)]=3-x` `⇔2x^2+5x-3=x^2-6x+9` `⇔x^2+11x-12=0` `⇔(x-1)(x+12)=0` \(⇔\left[ \begin{array}{l}x=1\\x=-12(l)\end{array} \right.\) Vậy `x=1` Bình luận
`ĐKXĐ: x\ge 1/2`
`2\sqrt(2x-1)+\sqrt(x+3)-\sqrt(5x+11)=0`
`⇔2\sqrt(2x-1)+\sqrt(x+3)=\sqrt(5x+11)`
`⇔4(2x-1)+x+3+4\sqrt[(2x-1)(x+3)]=5x+11`
`⇔4\sqrt[(2x-1)(x+3)]=12-4`
`⇔\sqrt[(2x-1)(x+3)]=3-x`
`⇔2x^2+5x-3=x^2-6x+9`
`⇔x^2+11x-12=0`
`⇔(x-1)(x+12)=0`
\(⇔\left[ \begin{array}{l}x=1\\x=-12(l)\end{array} \right.\)
Vậy `x=1`