Đáp án: $\begin{array}{l}1)\sqrt {16{x^2}} = 7\\ \Leftrightarrow 16{x^2} = 49\\ \Leftrightarrow {x^2} = \dfrac{{49}}{{16}}\\ \Leftrightarrow x = \pm \dfrac{7}{4}\\2)\sqrt {25{{\left( {x – 1} \right)}^2}} = 50\\ \Leftrightarrow 25{\left( {x – 1} \right)^2} = 2500\\ \Leftrightarrow {\left( {x – 1} \right)^2} = 100\\ \Leftrightarrow \left[ \begin{array}{l}x – 1 = 10\\x – 1 = – 10\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 11\\x = – 9\end{array} \right.\\3)\sqrt {4\left( {{x^2} – 4x + 4} \right)} = 12\\ \Leftrightarrow \sqrt {{2^2}{{\left( {x – 2} \right)}^2}} = 12\\ \Leftrightarrow \left[ \begin{array}{l}2\left( {x – 2} \right) = 12\\2\left( {x – 2} \right) = – 12\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x – 2 = 6\\x – 2 = – 6\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 8\\x = – 4\end{array} \right.\\4)\sqrt {4{x^2} + 4x + 1} – 6 = 0\\ \Leftrightarrow \sqrt {{{\left( {2x + 1} \right)}^2}} = 6\\ \Leftrightarrow \left| {2x + 1} \right| = 6\\ \Leftrightarrow \left[ \begin{array}{l}2x + 1 = 6\\2x + 1 = – 6\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{5}{2}\\x = – \dfrac{7}{2}\end{array} \right.\\5)\sqrt {5\left( {2x – 1} \right)} = 15\\ \Leftrightarrow 5\left( {2x – 1} \right) = 225\\ \Leftrightarrow 2x – 1 = 45\\ \Leftrightarrow x = 23\\6)\sqrt {3x – 5} + 4 = 7\\ \Leftrightarrow \sqrt {3x – 5} = 3\\ \Leftrightarrow 3x – 5 = 9\\ \Leftrightarrow x = \dfrac{{14}}{3}\\7)\sqrt {1 – 4x} – 3 = 7\\ \Leftrightarrow \sqrt {1 – 4x} = 10\\ \Leftrightarrow 1 – 4x = 100\\ \Leftrightarrow x = \dfrac{{ – 99}}{4}\\8)\sqrt {{x^2} – 6x + 9} – 9 = 0\\ \Leftrightarrow \sqrt {{{\left( {x – 3} \right)}^2}} = 9\\ \Leftrightarrow \left| {x – 3} \right| = 9\\ \Leftrightarrow \left[ \begin{array}{l}x – 3 = 9\\x – 3 = – 9\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 12\\x = – 6\end{array} \right.\\9)\sqrt { – 5x + 3} = 2\\ \Leftrightarrow – 5x + 3 = 4\\ \Leftrightarrow x = – \dfrac{1}{5}\\10)7 – \sqrt {x – 1} = 10\\ \Leftrightarrow \sqrt {x – 1} = – 3\left( {vo\,nghiem} \right)\end{array}$ Vậy pt vô nghiệm. Reply
Đáp án:
$\begin{array}{l}
1)\sqrt {16{x^2}} = 7\\
\Leftrightarrow 16{x^2} = 49\\
\Leftrightarrow {x^2} = \dfrac{{49}}{{16}}\\
\Leftrightarrow x = \pm \dfrac{7}{4}\\
2)\sqrt {25{{\left( {x – 1} \right)}^2}} = 50\\
\Leftrightarrow 25{\left( {x – 1} \right)^2} = 2500\\
\Leftrightarrow {\left( {x – 1} \right)^2} = 100\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 10\\
x – 1 = – 10
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 11\\
x = – 9
\end{array} \right.\\
3)\sqrt {4\left( {{x^2} – 4x + 4} \right)} = 12\\
\Leftrightarrow \sqrt {{2^2}{{\left( {x – 2} \right)}^2}} = 12\\
\Leftrightarrow \left[ \begin{array}{l}
2\left( {x – 2} \right) = 12\\
2\left( {x – 2} \right) = – 12
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 6\\
x – 2 = – 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = – 4
\end{array} \right.\\
4)\sqrt {4{x^2} + 4x + 1} – 6 = 0\\
\Leftrightarrow \sqrt {{{\left( {2x + 1} \right)}^2}} = 6\\
\Leftrightarrow \left| {2x + 1} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 6\\
2x + 1 = – 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = – \dfrac{7}{2}
\end{array} \right.\\
5)\sqrt {5\left( {2x – 1} \right)} = 15\\
\Leftrightarrow 5\left( {2x – 1} \right) = 225\\
\Leftrightarrow 2x – 1 = 45\\
\Leftrightarrow x = 23\\
6)\sqrt {3x – 5} + 4 = 7\\
\Leftrightarrow \sqrt {3x – 5} = 3\\
\Leftrightarrow 3x – 5 = 9\\
\Leftrightarrow x = \dfrac{{14}}{3}\\
7)\sqrt {1 – 4x} – 3 = 7\\
\Leftrightarrow \sqrt {1 – 4x} = 10\\
\Leftrightarrow 1 – 4x = 100\\
\Leftrightarrow x = \dfrac{{ – 99}}{4}\\
8)\sqrt {{x^2} – 6x + 9} – 9 = 0\\
\Leftrightarrow \sqrt {{{\left( {x – 3} \right)}^2}} = 9\\
\Leftrightarrow \left| {x – 3} \right| = 9\\
\Leftrightarrow \left[ \begin{array}{l}
x – 3 = 9\\
x – 3 = – 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 12\\
x = – 6
\end{array} \right.\\
9)\sqrt { – 5x + 3} = 2\\
\Leftrightarrow – 5x + 3 = 4\\
\Leftrightarrow x = – \dfrac{1}{5}\\
10)7 – \sqrt {x – 1} = 10\\
\Leftrightarrow \sqrt {x – 1} = – 3\left( {vo\,nghiem} \right)
\end{array}$
Vậy pt vô nghiệm.