`4x^2+3x+3=4x.\sqrt{x+3}+2.\sqrt{2x-1} `
`ĐKXĐ` : `x+3 ≥0`
` 2x-1 ≥ 0`
`⇔x≥1/2`
`4x^2+3x+3=4x.\sqrt{x+3}+2.\sqrt{2x-1} `
⇔`[(4x^2)-(4x.\sqrt{x+3})+(x+3)]+[(2x-1)+(2.\sqrt{2x-1})+1]=0`
⇔`(2x- \sqrt{x+3})^2+(\sqrt{2x-1}-1)^2=0`
⇒`(2x- \sqrt{x+3})^2=0`
` (\sqrt{2x-1}-1)^2=0 `
Xét :
`2x-\sqrt{x+3}=0`
⇔`4x^2-(x+3)=0`
⇔`4x^2-x-3=0`
⇔`4x^2+3x-4x-3=0`
⇔`x(4x+3)-(4x+3)=0`
⇔`(x-1)(4x+3)=0`
⇔\(\left[ \begin{array}{l}x-1=0\\4x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=\frac{-3}{4}(KTM)\end{array} \right.\)
⇔`x=1`
Xét :
`\sqrt{2x-1}-1=0`
`⇔2x-1-1=0`
`⇔2x-2=0`
`⇔x-1=0`
`⇔x=1 `
Vậy `x=1`
