Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\{x^2} – 2x + 2 = 2\sqrt {2x – 3} \,\,\,\,\,\left( {x \ge \dfrac{3}{2}} \right)\\ \Leftrightarrow {x^2} = 2x – 2 + 2\sqrt {2x – 3} \\ \Leftrightarrow {x^2} = \left( {2x – 3} \right) + 2.\sqrt {2x – 3} .1 + 1\\ \Leftrightarrow {x^2} = {\left( {\sqrt {2x – 3} + 1} \right)^2}\\ \Leftrightarrow \left[ \begin{array}{l}x = \sqrt {2x – 3} + 1\\x = – \sqrt {2x – 3} – 1\,\,\,\,\,\left( {L,x \ge \dfrac{3}{2}} \right)\end{array} \right.\\ \Leftrightarrow x = \sqrt {2x – 3} + 1\\ \Leftrightarrow x – 1 = \sqrt {2x – 3} \\ \Leftrightarrow {\left( {x – 1} \right)^2} = 2x – 3\,\,\,\,\,\,\left( {x \ge \dfrac{3}{2} \Rightarrow x > 1} \right)\\ \Leftrightarrow {x^2} – 2x + 1 = 2x – 3\\ \Leftrightarrow {x^2} – 4x + 4 = 0\\ \Leftrightarrow {\left( {x – 2} \right)^2} = 0\\ \Leftrightarrow x = 2\\b,\\{x^2} + 7x + 18 = 4.\sqrt {3x + 10} \,\,\,\,\,\,\,\,\,\left( {x \ge – \dfrac{{10}}{3}} \right)\\ \Leftrightarrow {x^2} + 7x + 18 – 4\sqrt {3x + 10} = 0\\ \Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left[ {3x + 14 – 4\sqrt {3x + 10} } \right] = 0\\ \Leftrightarrow {\left( {x + 2} \right)^2} + \left[ {\left( {3x + 10} \right) – 2.\sqrt {3x + 10} .2 + 4} \right] = 0\\ \Leftrightarrow {\left( {x + 2} \right)^2} + {\left( {\sqrt {3x + 10} – 2} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {x + 2} \right)^2} = 0\\{\left( {\sqrt {3x + 10} – 2} \right)^2} = 0\end{array} \right. \Leftrightarrow x = – 2\\c,\\{x^2} + 9x + 20 = 2\sqrt {3x + 10} \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge – \dfrac{{10}}{3}} \right)\\ \Leftrightarrow {x^2} + 9x + 20 – 2\sqrt {3x + 10} = 0\\ \Leftrightarrow \left( {{x^2} + 6x + 9} \right) + \left( {3x + 11 – 2\sqrt {3x + 10} } \right) = 0\\ \Leftrightarrow {\left( {x + 3} \right)^2} + \left[ {\left( {3x + 10} \right) – 2.\sqrt {3x + 10} .1 + 1} \right] = 0\\ \Leftrightarrow {\left( {x + 3} \right)^2} + {\left( {\sqrt {3x + 10} – 1} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {x + 3} \right)^2} = 0\\{\left( {\sqrt {3x + 10} – 1} \right)^2} = 0\end{array} \right. \Leftrightarrow x = – 3\\d,\\5x – 5 – 2\sqrt {2x – 5} = 4\sqrt {3x – 5} \,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{5}{2}} \right)\\ \Leftrightarrow 5x – 5 – 2\sqrt {2x – 5} – 4\sqrt {3x – 5} = 0\\ \Leftrightarrow \left( {2x – 4 – 2\sqrt {2x – 5} } \right) + \left( {3x – 1 – 4\sqrt {3x – 5} } \right) = 0\\ \Leftrightarrow \left[ {\left( {2x – 5} \right) – 2\sqrt {2x – 5} + 1} \right] + \left[ {\left( {3x – 5} \right) – 4\sqrt {3x – 5} + 4} \right] = 0\\ \Leftrightarrow {\left( {\sqrt {2x – 5} – 1} \right)^2} + {\left( {\sqrt {3x – 5} – 2} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {\sqrt {2x – 5} – 1} \right)^2} = 0\\{\left( {\sqrt {3x – 5} – 2} \right)^2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sqrt {2x – 5} = 1\\\sqrt {3x – 5} = 2\end{array} \right. \Leftrightarrow x = 3\\e,\\{x^2} – 3x – 2\sqrt {x – 1} + 4 = 0\,\,\,\,\,\,\left( {x \ge 1} \right)\\ \Leftrightarrow \left( {{x^2} – 4x + 4} \right) + \left( {x – 2\sqrt {x – 1} } \right) = 0\\ \Leftrightarrow {\left( {x – 2} \right)^2} + \left[ {\left( {x – 1} \right) – 2\sqrt {x – 1} + 1} \right] = 0\\ \Leftrightarrow {\left( {x – 2} \right)^2} + {\left( {\sqrt {x – 1} – 1} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {x – 2} \right)^2} = 0\\{\left( {\sqrt {x – 1} – 1} \right)^2} = 0\end{array} \right. \Leftrightarrow x = 2\\f,\\2{x^2} + 2x + 1 = \sqrt {4x + 1} \,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge – \dfrac{1}{4}} \right)\\ \Leftrightarrow 2{x^2} + 2x + 1 – \sqrt {4x + 1} = 0\\ \Leftrightarrow 4{x^2} + 4x + 2 – 2\sqrt {4x + 1} = 0\\ \Leftrightarrow 4{x^2} + \left[ {\left( {4x + 1} \right) – 2\sqrt {4x + 1} + 1} \right] = 0\\ \Leftrightarrow 4{x^2} + {\left( {\sqrt {4x + 1} – 1} \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}4{x^2} = 0\\{\left( {\sqrt {4x + 1} – 1} \right)^2} = 0\end{array} \right. \Leftrightarrow x = 0\end{array}\) Reply
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} – 2x + 2 = 2\sqrt {2x – 3} \,\,\,\,\,\left( {x \ge \dfrac{3}{2}} \right)\\
\Leftrightarrow {x^2} = 2x – 2 + 2\sqrt {2x – 3} \\
\Leftrightarrow {x^2} = \left( {2x – 3} \right) + 2.\sqrt {2x – 3} .1 + 1\\
\Leftrightarrow {x^2} = {\left( {\sqrt {2x – 3} + 1} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt {2x – 3} + 1\\
x = – \sqrt {2x – 3} – 1\,\,\,\,\,\left( {L,x \ge \dfrac{3}{2}} \right)
\end{array} \right.\\
\Leftrightarrow x = \sqrt {2x – 3} + 1\\
\Leftrightarrow x – 1 = \sqrt {2x – 3} \\
\Leftrightarrow {\left( {x – 1} \right)^2} = 2x – 3\,\,\,\,\,\,\left( {x \ge \dfrac{3}{2} \Rightarrow x > 1} \right)\\
\Leftrightarrow {x^2} – 2x + 1 = 2x – 3\\
\Leftrightarrow {x^2} – 4x + 4 = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} = 0\\
\Leftrightarrow x = 2\\
b,\\
{x^2} + 7x + 18 = 4.\sqrt {3x + 10} \,\,\,\,\,\,\,\,\,\left( {x \ge – \dfrac{{10}}{3}} \right)\\
\Leftrightarrow {x^2} + 7x + 18 – 4\sqrt {3x + 10} = 0\\
\Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left[ {3x + 14 – 4\sqrt {3x + 10} } \right] = 0\\
\Leftrightarrow {\left( {x + 2} \right)^2} + \left[ {\left( {3x + 10} \right) – 2.\sqrt {3x + 10} .2 + 4} \right] = 0\\
\Leftrightarrow {\left( {x + 2} \right)^2} + {\left( {\sqrt {3x + 10} – 2} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} = 0\\
{\left( {\sqrt {3x + 10} – 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = – 2\\
c,\\
{x^2} + 9x + 20 = 2\sqrt {3x + 10} \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge – \dfrac{{10}}{3}} \right)\\
\Leftrightarrow {x^2} + 9x + 20 – 2\sqrt {3x + 10} = 0\\
\Leftrightarrow \left( {{x^2} + 6x + 9} \right) + \left( {3x + 11 – 2\sqrt {3x + 10} } \right) = 0\\
\Leftrightarrow {\left( {x + 3} \right)^2} + \left[ {\left( {3x + 10} \right) – 2.\sqrt {3x + 10} .1 + 1} \right] = 0\\
\Leftrightarrow {\left( {x + 3} \right)^2} + {\left( {\sqrt {3x + 10} – 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 3} \right)^2} = 0\\
{\left( {\sqrt {3x + 10} – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = – 3\\
d,\\
5x – 5 – 2\sqrt {2x – 5} = 4\sqrt {3x – 5} \,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{5}{2}} \right)\\
\Leftrightarrow 5x – 5 – 2\sqrt {2x – 5} – 4\sqrt {3x – 5} = 0\\
\Leftrightarrow \left( {2x – 4 – 2\sqrt {2x – 5} } \right) + \left( {3x – 1 – 4\sqrt {3x – 5} } \right) = 0\\
\Leftrightarrow \left[ {\left( {2x – 5} \right) – 2\sqrt {2x – 5} + 1} \right] + \left[ {\left( {3x – 5} \right) – 4\sqrt {3x – 5} + 4} \right] = 0\\
\Leftrightarrow {\left( {\sqrt {2x – 5} – 1} \right)^2} + {\left( {\sqrt {3x – 5} – 2} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {\sqrt {2x – 5} – 1} \right)^2} = 0\\
{\left( {\sqrt {3x – 5} – 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {2x – 5} = 1\\
\sqrt {3x – 5} = 2
\end{array} \right. \Leftrightarrow x = 3\\
e,\\
{x^2} – 3x – 2\sqrt {x – 1} + 4 = 0\,\,\,\,\,\,\left( {x \ge 1} \right)\\
\Leftrightarrow \left( {{x^2} – 4x + 4} \right) + \left( {x – 2\sqrt {x – 1} } \right) = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} + \left[ {\left( {x – 1} \right) – 2\sqrt {x – 1} + 1} \right] = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} + {\left( {\sqrt {x – 1} – 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 2} \right)^2} = 0\\
{\left( {\sqrt {x – 1} – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = 2\\
f,\\
2{x^2} + 2x + 1 = \sqrt {4x + 1} \,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge – \dfrac{1}{4}} \right)\\
\Leftrightarrow 2{x^2} + 2x + 1 – \sqrt {4x + 1} = 0\\
\Leftrightarrow 4{x^2} + 4x + 2 – 2\sqrt {4x + 1} = 0\\
\Leftrightarrow 4{x^2} + \left[ {\left( {4x + 1} \right) – 2\sqrt {4x + 1} + 1} \right] = 0\\
\Leftrightarrow 4{x^2} + {\left( {\sqrt {4x + 1} – 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
4{x^2} = 0\\
{\left( {\sqrt {4x + 1} – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = 0
\end{array}\)