Giải phương trình 4x ² +3x +3 = 4x √(x+3) +2 √(2x-1) Giải hộ mình với
2 câu trả lời
`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`
Đáp án:
\(x=1.\)
Giải thích các bước giải: \[\begin{array}{l} \,\,\,\,\,\,\,4{x^2} + 3x + 3 = 4x\sqrt {x + 3} + 2\sqrt {2x - 1} \,\,\,\,\left( {DK:\,\,\,x \ge \frac{1}{2}} \right)\\ \Leftrightarrow 4{x^2} - 4x\sqrt {x + 3} + x + 3 + 2x - 1 - 2\sqrt {2x - 1} + 1 = 0\\ \Leftrightarrow {\left( {2x - \sqrt {x + 3} } \right)^2} + {\left( {\sqrt {2x - 1} - 1} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l} 2x - \sqrt {x + 3} = 0\\ \sqrt {2x - 1} - 1 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2x = \sqrt {x + 3} \\ \sqrt {2x - 1} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 4{x^2} = x + 3\\ 2x - 1 = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 4{x^2} - x - 3 = 0\\ 2x = 2 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x = 1\\ x = - \frac{3}{4}\,\,\,\left( {ktm} \right) \end{array} \right.\\ x = 1 \end{array} \right. \Leftrightarrow x = 1\,\,\left( {tm} \right). \end{array}\]