[toán 9] Các bài tập về căn bậc hai

Q

quynhsieunhan

1, Có: 8 < 9 \Rightarrow $\sqrt{8} < \sqrt{9}$ \Rightarrow $2\sqrt{2} < 3$
\Rightarrow $6 + 2\sqrt{2} < 6 + 3 = 9$
2. $\sqrt{3} - \sqrt{2} < \sqrt{4} - \sqrt{1}$
\Rightarrow $\frac{5}{\sqrt{3} - \sqrt{2}} < \frac{5}{\sqrt{4} - \sqrt{1}}$
\Rightarrow $\sqrt{3} + \sqrt{2} < \sqrt{4} + \sqrt{4} + 1 = 3$
3. $\sqrt{5} > 2$ \Rightarrow $2 + \sqrt{5} > 4$ \Rightarrow $(2 + \sqrt{5})^2 > 16$ \Leftrightarrow $9 + 4\sqrt{5} > 16$
 
S

soccan

1 d)
Có $ (\sqrt{11}-\sqrt{3})^2=14-2\sqrt{33} < 14-2\sqrt{25}=4$
hay $\sqrt{11}-\sqrt{3} <2$ (đpcm)
 
N

nhoc_surita

Bài 3: Rút gọn

a, $2\sqrt{2}(\sqrt{3} - 2) + (1 + 2\sqrt{2})^2 - 2\sqrt{6}$

b, $\frac{\sqrt{6} + \sqrt{14}}{2\sqrt{3} + \sqrt{28}}$

c, $\frac{\sqrt{2} + \sqrt{3} + \sqrt{6} + \sqrt{8} + \sqrt{16}}{\sqrt{2} + \sqrt{3} + \sqrt{4}}$
 
L

lp_qt

a. $\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}$

$=\dfrac{\sqrt{2}(\sqrt{3}+\sqrt{7})}{2(\sqrt{3}+\sqrt{7})}$

$=\dfrac{\sqrt{2}}{2}$

c. $\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

$=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}.(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

$=\sqrt{2}+1$
 
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