Phiếu bài tập tuần Toán 9 - Tuần 01

Số

121

144

169

225

256

324

361

400

0,01

CBH

CBHSH

$\sqrt{-2x+3}$

$\sqrt{-5x}$

$\sqrt{\dfrac{x}{3}}$

$\sqrt{1+{{x}^{2}}}$

$\sqrt{\dfrac{4}{x+3}}$

$\sqrt{\dfrac{-5}{{{x}^{2}}+6}}$

$\sqrt{\dfrac{1}{-1+x}}$

$\sqrt{\dfrac{2}{{{x}^{2}}}}$

$\sqrt{{{x}^{2}}-2x+1}$

$\sqrt{-{{x}^{2}}-2x-1}$

$\dfrac{1}{\sqrt{4{{x}^{2}}-12x+9}}$

$\sqrt{{{x}^{2}}-8x+15}$

$\sqrt{x-2}+\dfrac{1}{x-5}$

$\sqrt{\dfrac{2+x}{5-x}}$

$\sqrt{\dfrac{x-1}{x+2}}$

$\sqrt{{{(4-3\sqrt{2})}^{2}}}$

$\sqrt{{{(2+\sqrt{5})}^{2}}}$

$\sqrt{{{(4+\sqrt{2})}^{2}}}$

$\sqrt{6-2\sqrt{5}}$

$\sqrt{7+4\sqrt{3}}$

$\sqrt{12-6\sqrt{3}}$

$\sqrt{17+12\sqrt{2}}$

$\dfrac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}$

$\sqrt{6+2\sqrt{4-2\sqrt{3}}}$.

Số

121

144

169

225

256

324

361

400

0,01

CBH

11; -11

12 ;-12

13 ;-13

15; -15

14; -14

18; -18

19; -19

20; -20

0,1;-0,1

CBHSH

11

12

13

15

14

18

19

20

0,1

$x$

$4$

$\pm 0,3$

$-5$

$0$

$\pm 1$

$13$

$\pm 16$

$0,1$

$-0,1$

${{x}^{2}}$

$6$

$0,09$

$25$

$0$

$1$

$169$

$256$

$0,01$

$0,01$

$\sqrt{\left| x \right|}$

$2$

$\sqrt{0,3}$

$\sqrt{5}$

$0$

$1$

$\sqrt{13}$

$4$

$\sqrt{0,1}$

$\sqrt{0,1}$

$\sqrt{{{x}^{2}}}$

$4$

$0,3$

$5$

$0$

$1$

$13$

$16$

$0,1$

$0,1$

$-2\text{x}+3\ge 0\Leftrightarrow x\le \dfrac{3}{2}$

$-5\text{x}\ge 0\Leftrightarrow x\le 0$

$\dfrac{x}{3}\ge 0\Leftrightarrow x\ge 0$

$1+{{x}^{2}}\ge 0\Leftrightarrow x\in R$

$\begin{array}{l}
\left\{ \begin{array}{l}
\frac{4}{{x + 3}} \ge 0\\
x + 3 \ne 0
\end{array} \right.\\
 \Leftrightarrow x >  - 3
\end{array}$

$\dfrac{-5}{{{x}^{2}}+6}<0,\forall x$

$\Rightarrow$ $x\in \varnothing$

     $\begin{array}{l}
\left\{ \begin{array}{l}
\frac{1}{{ - 1 + x}} \ge 0\\
 - 1 + x \ne 0
\end{array} \right.\\
 \Leftrightarrow x > 1
\end{array}$

$\left\{ \begin{array}{l}
\frac{2}{{{x^2}}} \ge 0\\
{x^2} \ne 0
\end{array} \right. \Leftrightarrow x \ne 0$

$\begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0{\rm{ }}\forall x\\
 \Rightarrow x \in R
\end{array}$

$\begin{array}{l}
{\left( {x + 1} \right)^2} \le 0\\
 \Leftrightarrow x =  - 1
\end{array}$

${{\left( 2\text{x}-3 \right)}^{2}}>0\Leftrightarrow x\ne \dfrac{3}{2}$

$\begin{array}{l}
\left( {x - 5} \right).(x - 3) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \le 3\\
x \ge 5
\end{array} \right.
\end{array}$

$\left\{ \begin{array}{l}
x - 2 \ge 0\\
x - 5 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \ne 5
\end{array} \right.$

$\begin{array}{l}
\left\{ \begin{array}{l}
\frac{{2 + x}}{{5 - x}} \ge 0\\
5 - x \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - 2 \le x \le 5\\
x \ne 5
\end{array} \right.\\
 \Leftrightarrow  - 2 \le x < 5
\end{array}$

$\begin{array}{l}
\left\{ \begin{array}{l}
\frac{{x - 1}}{{x + 2}} \ge 0\\
x + 2 \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1\\
x \le  - 2
\end{array} \right.\\
x \ne  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x <  - 2
\end{array} \right.
\end{array}$