Phiếu bài tập tuần Toán 9 - Tuần 03

$\sqrt{\frac{121}{144}}$

$\sqrt{1\frac{17}{64}}$

$\sqrt{\frac{48}{75}}$

$\frac{\sqrt{192}}{\sqrt{12}}$

$\frac{\sqrt{{{6}^{5}}}}{\sqrt{{{2}^{3}}{{.3}^{5}}}}$

$\sqrt{\frac{0,99}{0,81}}$

$\sqrt{\frac{0,01}{0,0004}}$

$\frac{\sqrt{72}}{\sqrt{2}}$

$\sqrt{3,6.16,9}$

$\frac{\sqrt{12,5}}{\sqrt{0,5}}$

$\frac{\sqrt{a-2\sqrt{ab}+b}}{\sqrt{\sqrt{a}-\sqrt{b}}}$ với $a>b>0$

$\frac{\sqrt{x-3}}{\sqrt{\sqrt{x}+\sqrt{3}}}:\frac{\sqrt{\sqrt{x}-\sqrt{3}}}{\sqrt{3}}$ với $(x>3)$

$2{{y}^{2}}\sqrt{\frac{{{x}^{4}}}{4{{y}^{2}}}}$ với  $y<0;$

$\frac{y}{x}.\sqrt {\frac{{{x^2}}}{{{y^4}}}} $ với $x > 0;{\rm{ }}y \ne 0$

$5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} $ với $x < 0;{\rm{ }}y > 0$

$A=(3\sqrt{18}+2\sqrt{50}-4\sqrt{72}):8\sqrt{2}$

$B=(-4\sqrt{20}+5\sqrt{500}-3\sqrt{45}):\sqrt{5}$

$C=(\frac{\sqrt{3}+1}{\sqrt{3}-1}-\frac{\sqrt{3}-1}{\sqrt{3}+1}):\sqrt{48}$

a) ${{x}^{2}}\text{ }7$

b) ${{x}^{4}}-3$

c) ${{x}^{2}}\text{ }2\sqrt{13{{x}^{2}}}\text{ }+\text{ }13$

d) ${{x}^{2}}16$

e) $x-81$

f) ${{x}^{2}}+\text{ }2\sqrt{5}x\text{ }+\text{ }5$ 

$\sqrt{16x}=8$

$\sqrt{4x}=\sqrt{5}$

$\sqrt{2x-1}=\sqrt{5}$

$\sqrt{x-10}=-2$

$\sqrt{4({{x}^{2}}-2x+1)}-6=0$

$\sqrt{2}x-\sqrt{50}=0$

$\sqrt{4{{x}^{2}}}=\sqrt{x+5}$ (ĐK: $x+5\ge 0$ và bình phương 2 vế)

$\frac{11}{12}$

$\sqrt{\frac{81}{64}}=\frac{9}{8}$

$\sqrt{\frac{16}{25}}=\frac{4}{5}$

$\sqrt{16}=4$

$\frac{{\sqrt {{6^5}} }}{{\sqrt {{2^3}{{.3}^5}} }}$

$\sqrt{\frac{99}{81}}=\frac{\sqrt{11}}{3}$

$\sqrt{\frac{1}{0,04}}=\sqrt{25}=5$

$\sqrt{36}=6$

$\frac{{\sqrt {a - 2\sqrt {ab}  + b} }}{{\sqrt {\sqrt a  - \sqrt b } }}$ với $(a > b > 0)$

$\frac{{{y}^{2}}.{{x}^{2}}}{\left| y \right|}=-y.{{x}^{2}}$

với $y < 0;$

$\begin{array}{l}
\frac{y}{x}.\sqrt {\frac{{{x^2}}}{{{y^4}}}} \\
 = \frac{{y.\left| x \right|}}{{x.{y^2}}} = \frac{1}{y}
\end{array}$ với $x > 0;{\rm{ }}y \ne 0$

$\begin{array}{l}
A = (3\sqrt {18}  + 2\sqrt {50}  - 4\sqrt {72} ):8\sqrt 2 \\
 = \frac{{3\sqrt {18} }}{{8\sqrt 2 }} + \frac{{2\sqrt {50} }}{{8\sqrt 2 }} - \frac{{4\sqrt {72} }}{{8\sqrt 2 }}\\
 = \frac{9}{8} + \frac{{10}}{8} - \frac{{24}}{8} = \frac{{ - 5}}{8}
\end{array}$

$\begin{array}{l}
B = ( - 4\sqrt {20}  + 5\sqrt {500}  - 3\sqrt {45} ):\sqrt 5 \\
 =  - 4\sqrt 4  + 5\sqrt {100}  - 3\sqrt 9 \\
 =  - 8 + 50 - 9 = 33
\end{array}$

$\begin{array}{l}
C = \frac{{{{\left( {\sqrt 3  + 1} \right)}^2} - {{\left( {\sqrt 3  - 1} \right)}^2}}}{{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}}:4\sqrt 3 \\
 = \frac{{3 + 2\sqrt 3  + 1 - 3 + 2\sqrt 3  - 1}}{2}:4\sqrt 3 \\
 = \frac{{2\sqrt 3 }}{{4\sqrt 3 }} = \frac{1}{2}
\end{array}$

$\sqrt{16x}=8\Leftrightarrow 16x=64\Leftrightarrow x=4$

$\sqrt{4x}=\sqrt{5}\Leftrightarrow 4x=5\Leftrightarrow x=\frac{5}{4}$

$\sqrt{2x-1}=\sqrt{5}\Leftrightarrow 2x-1=5\Leftrightarrow x=3$

$\sqrt{x-10}=-2\Leftrightarrow x\in \varnothing $

$\begin{array}{l}
\sqrt {4({x^2} - 2x + 1)}  - 6 = 0 \Leftrightarrow \sqrt {4({x^2} - 2x + 1)}  = 6\\
 \Leftrightarrow \left| {x - 1} \right| = 3
\end{array}$

$ \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 2
\end{array} \right.$

$\sqrt{2}x-\sqrt{50}=0\Leftrightarrow \sqrt{2}x=\sqrt{50}\Leftrightarrow x=5$

$\sqrt {4{x^2}}  = \sqrt {x + 5}  \Leftrightarrow \left\{ \begin{array}{l}
4{x^2} = x + 5\\
x + 5 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = \frac{5}{4}\\
x =  - 1
\end{array} \right.\\
x \ge  - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{5}{4}\\
x =  - 1
\end{array} \right.$