Rút gọn biểu thức số chứa căn bậc hai SVIP

Hướng dẫn giải:

a) $A=\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}$

$=\sqrt{3}\Big[ \dfrac{(\sqrt{\sqrt{3}+1}+1)-(\sqrt{\sqrt{3}+1}-1)}{( \sqrt{\sqrt{3}+1}-1)(\sqrt{\sqrt{3}+1}+1)} \Big]$

$=\sqrt{3}.\dfrac{2}{\sqrt{3}+1-1}=2$.

b) $\dfrac{15}{\sqrt{6}+1}=\dfrac{15(\sqrt{6}-1)}{6-1}=3\sqrt{6}-3$;

$\dfrac{4}{\sqrt{6}-2}=4 + 2\sqrt{6}$;

$\dfrac{12}{3-\sqrt{6}}=12+4\sqrt{6}$.

Suy ra $B=\big(3\sqrt{6}-3+4 + 2\sqrt{6} - 12-4\sqrt{6}\big)(\sqrt6 + 11)$

$=(\sqrt6 + 11)(\sqrt6 - 11) = -115$.

c) $C=4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}$

$=4.2\sqrt{5}-3.5\sqrt{5}+5.3\sqrt{5}-3\sqrt{\dfrac{25}{5}}$

$=8\sqrt{5}-15\sqrt{5}-3\sqrt{5}+15\sqrt{5}=5\sqrt{5}$.

Hướng dẫn giải:

a) $\Big( \sqrt{\dfrac{4}{3}}+\sqrt{3} \Big) . \sqrt{6}$

$=\Big( \sqrt{\dfrac{4 . 3}{3^2}}+\sqrt{3} \Big)\sqrt{6}$

$=\Big( \dfrac{2\sqrt{3}}{3}+\sqrt{3} \Big)\sqrt{6}$

$=\dfrac{2\sqrt{3}. \sqrt{6}}{3}+\sqrt{3}. \sqrt{6}$

$=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}$.

b) $( 1-2\sqrt{5} )^2$

$=1-4\sqrt{5}+(2\sqrt{5})^2$

$=1-4\sqrt{5}+20$

$=21-4\sqrt{5}$.

c) $2\sqrt{3}-\sqrt{27}$

$=2\sqrt{3}-3\sqrt{3}$

$=-\sqrt{3}$.

d) $\sqrt{45}-\sqrt{20}+\sqrt{5}$

$=3\sqrt{5}-2\sqrt{5}+\sqrt{5}$

$=2\sqrt{5}$.

Hướng dẫn giải:

a) $2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}$

$=2\sqrt{\dfrac{2 . 3}{3^2}}-4\sqrt{\dfrac{3 . 2}{2^2}}$

$=2 . \dfrac{\sqrt{6}}{3}-2.\sqrt{6}$

$=-\dfrac{4\sqrt{6}}{3}$.

b) $\dfrac{5\sqrt{48}-3\sqrt{27}+2\sqrt{12}}{\sqrt{3}}$

$=\dfrac{5\sqrt{4^2. 3}-3\sqrt{3^2.3}+2\sqrt{2^2. 3}}{\sqrt{3}}$

$=\dfrac{20\sqrt{3}-9\sqrt{3}+4\sqrt{3}}{\sqrt{3}}$

$=\dfrac{15\sqrt{3}}{\sqrt{3}}=15$

c) $\dfrac{1}{3+2\sqrt{2}}+\dfrac{4\sqrt{2}-4}{2-\sqrt{2}}$

$=\dfrac{1. \big( 3-2\sqrt{2} \big)}{\big( 3+2\sqrt{2} \big)\big( 3-2\sqrt{2} \big)}+\dfrac{4( \sqrt{2}-1)}{\sqrt{2}( \sqrt{2}-1 )}$

$=\dfrac{3-2\sqrt{2}}{9-8}+\dfrac{4}{\sqrt{2}}$

$=3-2\sqrt{2}+2\sqrt{2}$

$=3$.