Giải PT ($\sqrt[ ]{x +3}$ – $\sqrt[ ]{x +1}$ ) ($x^{2}$ +$\sqrt[ ]{x^{2} + 4x + 3}$ ) = 2x 12/07/2021 Bởi Genesis Giải PT ($\sqrt[ ]{x +3}$ – $\sqrt[ ]{x +1}$ ) ($x^{2}$ +$\sqrt[ ]{x^{2} + 4x + 3}$ ) = 2x
Điều kiện xác định $x\ge -1$ $\begin{array}{l} \left( {\sqrt {x + 3} – \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\\ \Leftrightarrow \left( {\sqrt {x + 3} – \sqrt {x + 1} } \right)\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\\ \Leftrightarrow \left( {x + 3 – x – 1} \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\\ \Leftrightarrow {x^2} + \sqrt {{x^2} + 4x + 3} = x\sqrt {x + 3} + x\sqrt {x + 1} \\ \Leftrightarrow {x^2} – x\sqrt {x + 1} + \sqrt {\left( {x + 1} \right)\left( {x + 3} \right)} – x\sqrt {\left( {x + 3} \right)} = 0\\ \Leftrightarrow x\left( {x – \sqrt {x + 1} } \right) + \sqrt {x + 3} \left( {\sqrt {x + 1} – x} \right) = 0\\ \Leftrightarrow \left( {x – \sqrt {x + 1} } \right)\left( {x – \sqrt {x + 3} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \sqrt {x + 1} \\ x = \sqrt {x + 3} \end{array} \right. \Rightarrow \left[ \begin{array}{l} {x^2} – x – 1 = 0\\ {x^2} – x – 3 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 \pm \sqrt 5 }}{2}\\ x = \dfrac{{1 \pm \sqrt {13} }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt 5 }}{2}\\ x = \dfrac{{1 – \sqrt 5 }}{2}\\ x = \dfrac{{1 + \sqrt {13} }}{2} \end{array} \right.\\ \Rightarrow S = \left\{ {\dfrac{{1 + \sqrt 5 }}{2};\dfrac{{1 – \sqrt 5 }}{2};\dfrac{{1 + \sqrt {13} }}{2}} \right\} \end{array}$ Bình luận
Điều kiện xác định $x\ge -1$
$\begin{array}{l} \left( {\sqrt {x + 3} – \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\\ \Leftrightarrow \left( {\sqrt {x + 3} – \sqrt {x + 1} } \right)\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\\ \Leftrightarrow \left( {x + 3 – x – 1} \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\\ \Leftrightarrow {x^2} + \sqrt {{x^2} + 4x + 3} = x\sqrt {x + 3} + x\sqrt {x + 1} \\ \Leftrightarrow {x^2} – x\sqrt {x + 1} + \sqrt {\left( {x + 1} \right)\left( {x + 3} \right)} – x\sqrt {\left( {x + 3} \right)} = 0\\ \Leftrightarrow x\left( {x – \sqrt {x + 1} } \right) + \sqrt {x + 3} \left( {\sqrt {x + 1} – x} \right) = 0\\ \Leftrightarrow \left( {x – \sqrt {x + 1} } \right)\left( {x – \sqrt {x + 3} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \sqrt {x + 1} \\ x = \sqrt {x + 3} \end{array} \right. \Rightarrow \left[ \begin{array}{l} {x^2} – x – 1 = 0\\ {x^2} – x – 3 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 \pm \sqrt 5 }}{2}\\ x = \dfrac{{1 \pm \sqrt {13} }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt 5 }}{2}\\ x = \dfrac{{1 – \sqrt 5 }}{2}\\ x = \dfrac{{1 + \sqrt {13} }}{2} \end{array} \right.\\ \Rightarrow S = \left\{ {\dfrac{{1 + \sqrt 5 }}{2};\dfrac{{1 – \sqrt 5 }}{2};\dfrac{{1 + \sqrt {13} }}{2}} \right\} \end{array}$