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Giải giúp mình vs ạ mình cảm ơn
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Doris
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Nhừ hình nha
Đáp án:
$\begin{array}{l}
B1)a)2\sqrt {48} – 4\sqrt {27} + \sqrt {75} + \sqrt {12} \\
= 2.\sqrt {16.3} – 4.\sqrt {9.3} + \sqrt {25.3} + \sqrt {4.3} \\
= 2.4\sqrt 3 – 4.3\sqrt 3 + 5\sqrt 3 + 2\sqrt 3 \\
= 8\sqrt 3 – 12\sqrt 3 + 5\sqrt 3 + 2\sqrt 3 \\
= 3\sqrt 3 \\
b)\sqrt {{{\left( {3 – \sqrt 5 } \right)}^2}} – \sqrt {20} \\
= 3 – \sqrt 5 – 2\sqrt 5 \\
= 3 – 3\sqrt 5 \\
B2)\\
a)Dkxd:x \ge 0\\
\sqrt {45x} – 2\sqrt {20x} + 2\sqrt {80x} = 21\\
\Rightarrow 3\sqrt {5x} – 2.2\sqrt {5x} + 2.4\sqrt {5x} = 21\\
\Rightarrow 7\sqrt {5x} = 21\\
\Rightarrow \sqrt {5x} = 3\\
\Rightarrow 5x = 9\\
\Rightarrow x = \dfrac{9}{5}\left( {tmdk} \right)\\
Vay\,x = \dfrac{9}{5}\\
b)\sqrt {{x^2} – 10x + 25} = 4\\
\Rightarrow \sqrt {{{\left( {x – 5} \right)}^2}} = 4\\
\Rightarrow \left[ \begin{array}{l}
x – 5 = 4\\
x – 5 = – 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 9\\
x = 1
\end{array} \right.\\
Vay\,x = 1\,hoac\,x = 9\\
B3))\\
a)Dkxd:x > 0;x \ne 1\\
Q = \dfrac{{\sqrt x }}{{\sqrt x – 1}} – \dfrac{2}{{\sqrt x + 1}} – \dfrac{2}{{x – 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) – 2\left( {\sqrt x – 1} \right) – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x – 2\sqrt x + 2 – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x – \sqrt x }}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b)x = 9\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 3\\
\Rightarrow Q = \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{3}{{3 + 1}} = \dfrac{3}{4}
\end{array}$