giải pt: $\sqrt[]{3x-1}$ – $\sqrt[]{x+1}$ = 3$x^{2}$ – 2x -1 06/07/2021 Bởi Adalynn giải pt: $\sqrt[]{3x-1}$ – $\sqrt[]{x+1}$ = 3$x^{2}$ – 2x -1
Đáp án: `S={1}` Giải thích các bước giải: `\qquad \sqrt{3x-1}-\sqrt{x+1}=3x^2-2x-1` $(1)$ $ĐK: \begin{cases}3x-1\ge 0\\x+1\ge 0\end{cases}$ `<=>`$\begin{cases}x\ge \dfrac{1}{3}\\x\ge -1\end{cases}$`=>x\ge 1/3` `(1)<=>{(\sqrt{3x-1}-\sqrt{x+1})(\sqrt{3x-1}+\sqrt{x+1})}/{\sqrt{3x-1}+\sqrt{x+1}}=3x^2-3x+x-1` `<=>{3x-1-(x-1)}/{\sqrt{3x-1}+\sqrt{x+1}}=3x(x-1)+(x-1)` `<=>{2(x-1)}/{\sqrt{3x-1}+\sqrt{x+1}}=(x-1)(3x+1)` `<=>(x-1). [2/{\sqrt{3x-1}+\sqrt{x+1}}-(3x+1)]=0` `<=>`$\left[\begin{array}{l}x-1=0\\\dfrac{2}{\sqrt{3x-1}+\sqrt{x+1}}-(3x+1)=0\end{array}\right.$ `<=>`$\left[\begin{array}{l}x=1\ (thỏa\ đk)\\\dfrac{2}{\sqrt{3x-1}+\sqrt{x+1}}=3x+1\ (2)\end{array}\right.$ $\\$ Giải `(2)` Với mọi `x\ge 1/3` ta có: $\quad \begin{cases}3x-1\ge 0\\x+1\ge \dfrac{4}{3}\end{cases}$ `=>\sqrt{3x-1}+\sqrt{x+1}\ge 0+\sqrt{4/3}=2/{\sqrt{3}}` `=>2/{\sqrt{3x-1}+\sqrt{x+1}}\le 2 : 2/{\sqrt{3}}=\sqrt{3}<2` $\\$ `\qquad 3x+1\ge 3. 1/ 3 +1=2` `=>` Phương trình `(2)` có `VT<2; VP\ge 2` `=>(2)` vô nghiệm Vậy phương trình đã cho có tập nghiệm `S={1}` Bình luận
$\begin{array}{l} ĐKXD:x \ge \dfrac{1}{3}\\ \sqrt {3x – 1} – \sqrt {x + 1} = 3{x^2} – 2x – 1\\ \Leftrightarrow 3{x^2} – 2x – 1 + \sqrt {x + 1} – \sqrt {3x – 1} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) + \dfrac{{x + 1 – 3x + 1}}{{\sqrt {3x – 1} + \sqrt {x + 1} }} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) + \dfrac{{2\left( {1 – x} \right)}}{{\sqrt {3x – 1} + \sqrt {x + 1} }} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1 – \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\\ 3x + 1 = \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }}\left( * \right) \end{array} \right.\\ \left( * \right) + 3x + 1 \ge 3.\dfrac{1}{3} + 1 = 2\\ \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }} \le \dfrac{2}{{\sqrt {\dfrac{4}{3}} }} = \sqrt 3 \\ \Rightarrow 3x + 1 – \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }} > 0\\ \left( * \right)VN\\ \Rightarrow S = \left\{ 1 \right\}\\ \end{array}$ Bình luận
Đáp án:
`S={1}`
Giải thích các bước giải:
`\qquad \sqrt{3x-1}-\sqrt{x+1}=3x^2-2x-1` $(1)$
$ĐK: \begin{cases}3x-1\ge 0\\x+1\ge 0\end{cases}$
`<=>`$\begin{cases}x\ge \dfrac{1}{3}\\x\ge -1\end{cases}$`=>x\ge 1/3`
`(1)<=>{(\sqrt{3x-1}-\sqrt{x+1})(\sqrt{3x-1}+\sqrt{x+1})}/{\sqrt{3x-1}+\sqrt{x+1}}=3x^2-3x+x-1`
`<=>{3x-1-(x-1)}/{\sqrt{3x-1}+\sqrt{x+1}}=3x(x-1)+(x-1)`
`<=>{2(x-1)}/{\sqrt{3x-1}+\sqrt{x+1}}=(x-1)(3x+1)`
`<=>(x-1). [2/{\sqrt{3x-1}+\sqrt{x+1}}-(3x+1)]=0`
`<=>`$\left[\begin{array}{l}x-1=0\\\dfrac{2}{\sqrt{3x-1}+\sqrt{x+1}}-(3x+1)=0\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=1\ (thỏa\ đk)\\\dfrac{2}{\sqrt{3x-1}+\sqrt{x+1}}=3x+1\ (2)\end{array}\right.$
$\\$
Giải `(2)`
Với mọi `x\ge 1/3` ta có:
$\quad \begin{cases}3x-1\ge 0\\x+1\ge \dfrac{4}{3}\end{cases}$
`=>\sqrt{3x-1}+\sqrt{x+1}\ge 0+\sqrt{4/3}=2/{\sqrt{3}}`
`=>2/{\sqrt{3x-1}+\sqrt{x+1}}\le 2 : 2/{\sqrt{3}}=\sqrt{3}<2`
$\\$
`\qquad 3x+1\ge 3. 1/ 3 +1=2`
`=>` Phương trình `(2)` có `VT<2; VP\ge 2`
`=>(2)` vô nghiệm
Vậy phương trình đã cho có tập nghiệm `S={1}`
$\begin{array}{l} ĐKXD:x \ge \dfrac{1}{3}\\ \sqrt {3x – 1} – \sqrt {x + 1} = 3{x^2} – 2x – 1\\ \Leftrightarrow 3{x^2} – 2x – 1 + \sqrt {x + 1} – \sqrt {3x – 1} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) + \dfrac{{x + 1 – 3x + 1}}{{\sqrt {3x – 1} + \sqrt {x + 1} }} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) + \dfrac{{2\left( {1 – x} \right)}}{{\sqrt {3x – 1} + \sqrt {x + 1} }} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {3x + 1 – \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\\ 3x + 1 = \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }}\left( * \right) \end{array} \right.\\ \left( * \right) + 3x + 1 \ge 3.\dfrac{1}{3} + 1 = 2\\ \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }} \le \dfrac{2}{{\sqrt {\dfrac{4}{3}} }} = \sqrt 3 \\ \Rightarrow 3x + 1 – \dfrac{2}{{\sqrt {3x – 1} + \sqrt {x + 1} }} > 0\\ \left( * \right)VN\\ \Rightarrow S = \left\{ 1 \right\}\\ \end{array}$