Toán Căn X^2-1 – căn 3x^2+4x+1 = (8-2x).căn x+1 16/07/2021 By Vivian Căn X^2-1 – căn 3x^2+4x+1 = (8-2x).căn x+1
`\sqrt(x^2-1)-\sqrt(3x^2+4x+1)=(8-2x)\sqrt(x+1)` `⇔\sqrt((x-1)(x+1))-\sqrt(3(x+1/3)(x+1))-(8-2x)\sqrt(x+1)=0` `⇔\sqrt(x+1)(\sqrt(x-1)-\sqrt(3(x+1/3))-8+2x)=0` ⇔\(\left[ \begin{array}{l}\sqrt(x+1)=0\\\sqrt(x-1)-3\sqrt(x+1/3)-8+2x=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x+1=0\\(x-5)(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4)=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-1\\x-5=0(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4>0)\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\) Trả lời
$\begin{array}{l} \sqrt {{x^2} – 1} – \sqrt {3{x^2} + 4x + 1} = \left( {8 – 2x} \right)\sqrt {x + 1} \\ \Leftrightarrow \sqrt {x – 1} .\sqrt {x + 1} – \sqrt {\left( {x + 1} \right)\left( {3x + 1} \right)} = 2\left( {4 – x} \right)\sqrt {x + 1} \\ \Leftrightarrow \sqrt {x + 1} \left( {\sqrt {x – 1} – \sqrt {3x + 1} + 2x – 8} \right) = 0\\ \Leftrightarrow \sqrt {x + 1} \left[ {\left( {\sqrt {x – 1} – 2} \right) – \left( {\sqrt {3x + 1} – 4} \right) + 2x – 10} \right] = 0\\ \Leftrightarrow \sqrt {x + 1} \left[ {\dfrac{{x – 5}}{{\sqrt {x + 1} + 2}} – \dfrac{{x – 5}}{{\sqrt {3x + 1} + 4}} + 2\left( {x – 5} \right)} \right] = 0\\ \Leftrightarrow \sqrt {x + 1} \left( {x – 5} \right)\left( {\dfrac{1}{{\sqrt {x + 1 + 2} }} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt {x + 1} = 0\\ x – 5 = 0\\ \dfrac{1}{{\sqrt {x + 1} + 2}} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2 = 0\left( * \right) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = – 1\\ x = 5\\ \dfrac{1}{{\sqrt {x + 1} + 2}} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2 = 0\left( * \right) \end{array} \right.\\ \left( * \right):\dfrac{1}{{\sqrt {3x + 1} + 4}} \le \dfrac{1}{4} \Rightarrow \dfrac{1}{{\sqrt {x + 1 + 2} }} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2 > 0\\ \Rightarrow S\left\{ { – 1;5} \right\}\\ \end{array}$ Trả lời
`\sqrt(x^2-1)-\sqrt(3x^2+4x+1)=(8-2x)\sqrt(x+1)`
`⇔\sqrt((x-1)(x+1))-\sqrt(3(x+1/3)(x+1))-(8-2x)\sqrt(x+1)=0`
`⇔\sqrt(x+1)(\sqrt(x-1)-\sqrt(3(x+1/3))-8+2x)=0`
⇔\(\left[ \begin{array}{l}\sqrt(x+1)=0\\\sqrt(x-1)-3\sqrt(x+1/3)-8+2x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x+1=0\\(x-5)(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4)=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x-5=0(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4>0)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)
$\begin{array}{l} \sqrt {{x^2} – 1} – \sqrt {3{x^2} + 4x + 1} = \left( {8 – 2x} \right)\sqrt {x + 1} \\ \Leftrightarrow \sqrt {x – 1} .\sqrt {x + 1} – \sqrt {\left( {x + 1} \right)\left( {3x + 1} \right)} = 2\left( {4 – x} \right)\sqrt {x + 1} \\ \Leftrightarrow \sqrt {x + 1} \left( {\sqrt {x – 1} – \sqrt {3x + 1} + 2x – 8} \right) = 0\\ \Leftrightarrow \sqrt {x + 1} \left[ {\left( {\sqrt {x – 1} – 2} \right) – \left( {\sqrt {3x + 1} – 4} \right) + 2x – 10} \right] = 0\\ \Leftrightarrow \sqrt {x + 1} \left[ {\dfrac{{x – 5}}{{\sqrt {x + 1} + 2}} – \dfrac{{x – 5}}{{\sqrt {3x + 1} + 4}} + 2\left( {x – 5} \right)} \right] = 0\\ \Leftrightarrow \sqrt {x + 1} \left( {x – 5} \right)\left( {\dfrac{1}{{\sqrt {x + 1 + 2} }} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt {x + 1} = 0\\ x – 5 = 0\\ \dfrac{1}{{\sqrt {x + 1} + 2}} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2 = 0\left( * \right) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = – 1\\ x = 5\\ \dfrac{1}{{\sqrt {x + 1} + 2}} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2 = 0\left( * \right) \end{array} \right.\\ \left( * \right):\dfrac{1}{{\sqrt {3x + 1} + 4}} \le \dfrac{1}{4} \Rightarrow \dfrac{1}{{\sqrt {x + 1 + 2} }} – \dfrac{1}{{\sqrt {3x + 1} + 4}} + 2 > 0\\ \Rightarrow S\left\{ { – 1;5} \right\}\\ \end{array}$