Bài tập theo tuần Toán 9 - Tuần 8 - Đề 1

Chọn đáp án A

Chọn đáp án C

Chọn đáp án B

Chọn đáp án A

$\sqrt{\frac{-1}{-2+x}}$ có nghĩa khi $-2+x<0\iff x<2$

$\begin{array}{l}6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}-6\sqrt{3}\\ =6.2\sqrt{3}-2\sqrt{5}-2.3\sqrt{3}+5\sqrt{5}-6\sqrt{3}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}-6\sqrt{3}=3\sqrt{5}\end{array}$

$\begin{array}{l}a) A=\left(\frac{x\sqrt{x}}{\sqrt{x}+1}+\frac{x^2}{x\sqrt{x}+x}\right).\left(1-\frac{1}{\sqrt{x}}\right)\left(x>0\right)\\ =\left(\frac{x\sqrt{x}}{\sqrt{x}+1}+\frac{x}{\sqrt{x}+1}\right).\frac{\sqrt{x}-1}{\sqrt{x}}\\ =\frac{x\left(\sqrt{x}+1\right)}{\sqrt{x}+1}.\frac{\sqrt{x}-1}{\sqrt{x}}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\end{array}$

b) 

$A=x-\sqrt{x}={\left(\sqrt{x}\right)}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}$

Vì ${\left(\sqrt{x}-\frac{1}{2}\right)}^2\ge 0\left(\forall x\right)\Rightarrow {\left(\sqrt{x}-\frac{1}{2}\right)}^2-\frac{1}{4}\ge -\frac{1}{4}\Rightarrow MinA=-\frac{1}{4}\iff x=\frac{1}{4}$

$\begin{array}{l}\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\\ =\frac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{\frac{33}{11}}+5\sqrt{\frac{4}{3}}\\ =2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{5.2\sqrt{3}}{3}=-9\sqrt{3}+\frac{10}{3}\sqrt{3}=-\frac{17}{3}\sqrt{3}\end{array}$