Bài tập: Khử mẫu biểu thức lấy căn SVIP

Hướng dẫn giải:

a) $\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{3.2}{2.2}}=\dfrac{\sqrt{6}}{\sqrt{2^{2}}}=\dfrac{\sqrt{6}}{2}$.

b) $\sqrt{\dfrac{3 a}{5 b}}=\sqrt{\dfrac{3 a \cdot 5 b}{5 b \cdot 5 b}}=\dfrac{\sqrt{15 a b}}{\sqrt{(5 b)^{2}}}=\dfrac{\sqrt{15 a b}}{5|b|}$.

c) $\sqrt{\dfrac{5}{12}}=\sqrt{\dfrac{5.12}{12.12}}=\dfrac{\sqrt{60}}{\sqrt{12^{2}}}=\dfrac{\sqrt{4.15}}{12}=\dfrac{2 \sqrt{15}}{12}=\dfrac{\sqrt{15}}{6}$.

d) $\sqrt{\dfrac{5 x}{18 y}}=\sqrt{\dfrac{5 x .18 y}{18 y .18 y}}=\dfrac{\sqrt{90 x y}}{\sqrt{(18 y)^{2}}}=\dfrac{3 \sqrt{10 x y}}{18|y|}=\dfrac{\sqrt{10 x y}}{6|y|}$.

e) $\sqrt{\dfrac{(1+\sqrt{2})^{3}}{27}}=\sqrt{\dfrac{(1+\sqrt{2})^{2} 3(1+\sqrt{2})}{9^{2}}}=\dfrac{1+\sqrt{2}}{9} \sqrt{3+3 \sqrt{2}} .$

Hướng dẫn giải:

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

b) $\dfrac{x}{y} \sqrt{\dfrac{x}{y}}=\dfrac{x}{y} \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x}{y^{2}} \sqrt{x y} .$

c) $\sqrt{\dfrac{1}{a}+\dfrac{1}{a^{2}}}=\sqrt{\dfrac{a+1}{a^{2}}}=\dfrac{\sqrt{a+1}}{a} .$

d) $\sqrt{\dfrac{4 x^{3}}{25 y}}=\sqrt{\dfrac{4 x^{2} x y}{25 y^{2}}}=\dfrac{2 x}{5 y} \sqrt{x y} .$

e) $2 a b \sqrt{\dfrac{3}{a b}}=2\sqrt{\dfrac{3(ab)^2}{ab}}=2\sqrt{3ab}$.

Hướng dẫn giải:

a) $A=\dfrac{\sqrt{6}}{3}+\dfrac{2 \sqrt{6}}{2}-\sqrt{6}=\dfrac{\sqrt{6}}{3}$.

b) $B=\dfrac{3}{5} \sqrt{10}+\dfrac{1}{2} \sqrt{10}-2 \sqrt{10}=-\dfrac{1}{2} \sqrt{10}$.

c) $C=-\dfrac{1}{5} \sqrt{15}+\dfrac{3}{3} \sqrt{15}-4 \sqrt{15}=-\dfrac{16}{5} \sqrt{15}$.

Hướng dẫn giải:

a) $M=\dfrac{1}{7} \sqrt{21 a}-\dfrac{2}{3} \sqrt{21 a}+\sqrt{21 a}=-\dfrac{10}{21} \sqrt{21 a}$.

b) $N=\dfrac{2}{3} \sqrt{6 x}-\dfrac{3}{2} \sqrt{6 x}+\sqrt{6 x}=\dfrac{1}{6} \sqrt{6 x}$.

c) $P=\dfrac{4}{5} \sqrt{10 y}+\dfrac{3}{2} \sqrt{10 y}-\sqrt{10 y}=\dfrac{13}{10} \sqrt{10 y}$.

Hướng dẫn giải:

a) Vì $\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1.5}{5.5}}=\dfrac{\sqrt{5}}{\sqrt{5^{2}}}=\dfrac{\sqrt{5}}{5} \cdot \sqrt{\dfrac{4}{5}}=\sqrt{\dfrac{4.5}{5.5}}=\dfrac{\sqrt{4.5}}{\sqrt{5^{2}}}=\dfrac{2 \sqrt{5}}{5}$, nên

$M=5 \cdot \dfrac{\sqrt{5}}{5}+\dfrac{5}{2} \cdot \dfrac{2 \sqrt{5}}{5}-3 \sqrt{5}=\sqrt{5}+\sqrt{5}-3 \sqrt{5}=-\sqrt{5}$

b) Vì $\quad \sqrt{\dfrac{1}{2}}=\sqrt{\dfrac{1.2}{2.2}}=\dfrac{\sqrt{2}}{\sqrt{2^{2}}}=\dfrac{\sqrt{2}}{2}, \sqrt{4.5}=\sqrt{\dfrac{9}{2}}=\dfrac{\sqrt{9.2}}{\sqrt{2^{2}}}=\dfrac{3 \sqrt{2}}{2}$

$\sqrt{12.5}=\sqrt{\dfrac{25}{2}}=\dfrac{\sqrt{25.2}}{\sqrt{2^{2}}}=\dfrac{5 \sqrt{2}}{2} \text { nên }$

$N=3 \cdot \dfrac{\sqrt{2}}{2}+\dfrac{3 \sqrt{2}}{2}-\dfrac{5 \sqrt{2}}{2}=\dfrac{\sqrt{2}}{2} .$

c) Vì $\quad \sqrt{\dfrac{1}{3}}=\sqrt{\dfrac{1.3}{3.3}}=\dfrac{\sqrt{3}}{\sqrt{3^{2}}}=\dfrac{\sqrt{3}}{3}, \sqrt{1 \dfrac{1}{3}}=\sqrt{\dfrac{4.3}{3.3}}=\dfrac{\sqrt{4.3}}{\sqrt{3^{2}}}=\dfrac{2 \sqrt{3}}{3}$ nên

$P=\dfrac{\sqrt{3}}{3}+\dfrac{2 \sqrt{3}}{3}+4 \sqrt{3}=\sqrt{3}\left(\dfrac{1}{3}+\dfrac{2}{3}+4\right)=5 \sqrt{3} \text {. }$

d) Vì $\quad \sqrt{\dfrac{4}{a}}=\sqrt{\dfrac{4 \cdot a}{a \cdot a}}=\dfrac{\sqrt{4 a}}{\sqrt{a^{2}}}=\dfrac{2 \sqrt{a}}{a} \cdot \sqrt{\dfrac{9 a}{a^{3}}}=\sqrt{\dfrac{9 \cdot a}{a^{3} \cdot a}}=\dfrac{3 \sqrt{a}}{\sqrt{\left(a^{2}\right)^{2}}}=\dfrac{3 \sqrt{a}}{a^{2}}$ nên

$Q=2 \sqrt{a}-a \cdot \dfrac{2 \sqrt{a}}{a}+a^{2} \cdot \dfrac{3 \sqrt{a}}{a^{2}}=2 \sqrt{a}-2 \sqrt{a}+3 \sqrt{a}=3 \sqrt{a} \text {. }$