Bài tập: Trục căn thức ở mẫu SVIP

Hướng dẫn giải:

a) $\dfrac{a}{\sqrt{a}}=\dfrac{(\sqrt{a})^{2}}{\sqrt{a}}=\sqrt{a}$.

b) $\dfrac{a}{\sqrt{a b}}=\dfrac{a \sqrt{a b}}{\sqrt{a b} \sqrt{a b}}=\dfrac{a \sqrt{a b}}{(\sqrt{a b})^{2}}=\dfrac{a \sqrt{a b}}{a b}=\dfrac{\sqrt{a b}}{b}$.

c) $\dfrac{x}{\sqrt{3 x^{3}}}=\dfrac{x \sqrt{3 x}}{\sqrt{3 x^{3}} \sqrt{3 x}}=\dfrac{x \sqrt{3 x}}{\left(\sqrt{3 x^{2}}\right)^{2}}=\dfrac{x \sqrt{3 x}}{3 x^{2}}=\dfrac{\sqrt{3 x}}{3 x}$.

d) $\dfrac{4 y^{2}}{\sqrt{2 y^{5}}}=\dfrac{4 y^{2} \sqrt{2 y}}{\sqrt{2 y^{5}} \sqrt{2 y}}=\dfrac{4 y^{2} \sqrt{2 y}}{\left(\sqrt{2 y^{3}}\right)^{2}}=\dfrac{4 y^{2} \sqrt{2 y}}{2 y^{3}}=\dfrac{2 \sqrt{2 y}}{y}$.

Hướng dẫn giải:

a) $\dfrac{1}{\sqrt{a}+b}=\dfrac{1(\sqrt{a}-b)}{(\sqrt{a}+b)(\sqrt{a}-b)}=\dfrac{\sqrt{a}-b}{(\sqrt{a})^{2}-b^{2}}=\dfrac{\sqrt{a}-b}{a-b^{2}}$.

b) $\dfrac{2}{\sqrt{a}-\sqrt{b}}=\dfrac{2(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}=\dfrac{2(\sqrt{a}+\sqrt{b})}{(\sqrt{a})^{2}-(\sqrt{b})^{2}}=\dfrac{2(\sqrt{a}+\sqrt{b})}{a-b}$.

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

d) $\dfrac{2 x y}{2 \sqrt{x}+3 \sqrt{y}}=\dfrac{2 x y(2 \sqrt{x}-3 \sqrt{y})}{(2 \sqrt{x}+3 \sqrt{y})(2 \sqrt{x}-3 \sqrt{y})}$

$=\dfrac{2 x y(2 \sqrt{x}-3 \sqrt{y})}{(2 \sqrt{x})^{2}-(3 \sqrt{y})^{2}}=\dfrac{2 x y(2 \sqrt{x}-3 \sqrt{y})}{4 x-9 y} .$

Hướng dẫn giải:

a) Vì $\dfrac{6}{\sqrt{7}-2}=\dfrac{6(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}=\dfrac{6(\sqrt{7}+2)}{7-4}=2 \sqrt{7}+4$ $\dfrac{5}{4+\sqrt{7}}=\dfrac{5(4-\sqrt{7})}{(4+\sqrt{7})(4-\sqrt{7})}=\dfrac{5(4-\sqrt{7})}{16-7}=\dfrac{20-5 \sqrt{7}}{9}$ nên A $=\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2 \sqrt{7}}{4}+2 \sqrt{7}+4+\dfrac{20-5 \sqrt{7}}{9}=\dfrac{32 \sqrt{7}-20}{9}$

b) $B=\dfrac{17 \sqrt{6}}{6}$.

Hướng dẫn giải:

a) $C=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{6}-(\sqrt{3}+\sqrt{2}-\sqrt{6})}{(\sqrt{3}+\sqrt{2}+\sqrt{6})(\sqrt{3}+\sqrt{2}-\sqrt{6})}=\dfrac{2 \sqrt{6}}{(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{6})^{2}}$

$=\dfrac{2 \sqrt{6}(2 \sqrt{6}+1)}{(2 \sqrt{6}-1)(2 \sqrt{6}+1)}=\dfrac{2 \sqrt{6}(2 \sqrt{6}+1)}{23} .$

b) $D=\left[\dfrac{-\sqrt{2}(1-\sqrt{3})}{1(1-\sqrt{3})}-\dfrac{\sqrt{5} \cdot \sqrt{5}}{\sqrt{5}}\right] \cdot(\sqrt{5}-\sqrt{2})=-(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})$

Hướng dẫn giải:

a) $E=2 \sqrt{80 \sqrt{3}}+3 \sqrt{20 \sqrt{3}}-2 \sqrt{5 \sqrt{3}}-12 \sqrt{5 \sqrt{3}};$

$=8 \sqrt{5 \sqrt{3}}+6 \sqrt{5 \sqrt{3}}-2 \sqrt{5 \sqrt{3}}-12 \sqrt{5 \sqrt{3}}=0.$

b) Vì $ \dfrac{5}{12}-\dfrac{1}{\sqrt{6}}=\dfrac{5-2 \sqrt{6}}{12}=\dfrac{(\sqrt{3}-\sqrt{2})^{2}}{12}.$

$\dfrac{1}{\sqrt{3}}+\dfrac{1}{2 \sqrt{3}}=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}=\dfrac{2 \sqrt{3}+\sqrt{2}}{6}.$

nên $F=\dfrac{2 \sqrt{3}+\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}} \sqrt{\dfrac{(\sqrt{3}-\sqrt{2})^{2}}{12}}=\dfrac{2 \sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3 \sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}.$