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a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)
\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)
b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)
c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)
a: \(=-xy\cdot\dfrac{\sqrt{xy}}{x}=-y\sqrt{yx}\)
b: \(=\sqrt{\dfrac{-105x^3}{35^2}}=\sqrt{-105x}\cdot\dfrac{x}{35}\)
c: \(=\sqrt{\dfrac{5a^3b}{49b^2}}=\sqrt{5ab}\cdot\dfrac{a}{7b}\)
d: \(=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3}\cdot\sqrt{xy}\)
a: \(A=\dfrac{\sqrt{6}}{3}+\sqrt{6}-\sqrt{6}=\dfrac{\sqrt{6}}{3}\)
b: \(B=\dfrac{3}{5}\sqrt{10}+\dfrac{1}{2}\sqrt{10}-2\sqrt{10}=-\dfrac{9}{10}\sqrt{10}\)
c: \(C=\dfrac{\sqrt{21}}{7}\cdot\sqrt{a}-2\cdot\dfrac{\sqrt{21}}{3}\cdot\sqrt{a}+\sqrt{21}\cdot\sqrt{a}\)
\(=\dfrac{10\sqrt{21a}}{21}\)
bài 1) a) \(xy\sqrt{\dfrac{x}{y}}=x\sqrt{y}\sqrt{y}\dfrac{\sqrt{x}}{\sqrt{y}}=x\sqrt{x}\sqrt{y}=\left(\sqrt{x}\right)^3\sqrt{y}\)
b) \(\sqrt{\dfrac{5a^3}{49b}}=\dfrac{\sqrt{5a^3}}{\sqrt{49b}}=\dfrac{\sqrt{5a^3}}{7\sqrt{b}}=\dfrac{\sqrt{5a^3}.\sqrt{b}}{7\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{5a^3b}}{7b}\)
bài 2) a) \(\dfrac{\sqrt{3}-3}{1-\sqrt{3}}=\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=\sqrt{3}\)
b) \(\dfrac{5-\sqrt{15}}{\sqrt{3}-\sqrt{5}}=\dfrac{-\sqrt{5}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}-\sqrt{5}}=-\sqrt{5}\)
c) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
a) \(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{1}}{10\sqrt{6}}=\dfrac{\sqrt{1}.\sqrt{6}}{10\sqrt{6}.\sqrt{6}}=\dfrac{\sqrt{6}}{60}\)
b) \(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{11}}{6\sqrt{15}}=\dfrac{\sqrt{11}.\sqrt{15}}{6\sqrt{15}.\sqrt{15}}=\dfrac{\sqrt{165}}{90}\)
c) \(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{3}}{5\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{5\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{10}\)
d) \(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{5}}{7\sqrt{2}}=\dfrac{\sqrt{5}.\sqrt{2}}{7\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}}{14}\)
e) \(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{\sqrt{\left(1-\sqrt{3}\right)^2}}{3\sqrt{3}}=\dfrac{\sqrt{3}-1}{3\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{3\sqrt{3}.\sqrt{3}}=\dfrac{3-\sqrt{3}}{9}\)
\(\sqrt{\dfrac{1}{600}}=\sqrt{\dfrac{1\cdot6}{600\cdot6}}=\sqrt{\dfrac{6}{60^2}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\sqrt{\dfrac{11\cdot15}{540\cdot15}}=\sqrt{\dfrac{165}{90^2}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\sqrt{\dfrac{3\cdot2}{50\cdot2}}=\sqrt{\dfrac{6}{10^2}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\sqrt{\dfrac{5\cdot2}{98\cdot2}}=\sqrt{\dfrac{10}{12^2}}=\dfrac{\sqrt{10}}{12}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\sqrt{\dfrac{3\left(1-\sqrt{3}\right)^2}{27\cdot3}}\)
\(=\dfrac{\sqrt{3\left(1-\sqrt{3}\right)^2}}{\sqrt{9^2}}=\dfrac{\left|1-\sqrt{3}\right|\cdot\sqrt{3}}{9}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\sqrt{3}}{9}\)
a. \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2.3}{3^2}}=\dfrac{1}{3}.\sqrt{6}\)
b. \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{5^2}}=\dfrac{x}{5}.\sqrt{5}\) (vì x \(\ge\) 0)
c. \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3.x}{x^2}}=\dfrac{1}{x}.\sqrt{3x}\) (vì x > 0)
d. \(\sqrt{x^2-\dfrac{x^2}{7}}=\sqrt{\dfrac{6x^2}{7}}=\sqrt{\dfrac{6x^2.7}{7.7}}=\sqrt{\dfrac{42.x^2}{7^2}}=-\dfrac{x}{7}.\sqrt{42}\) (vì x < 0)
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
có nghĩa khi
và
Nếuthì
Nếuthì
- Tương tự như vậy ta có:
Nếuthì
Nếu - Ta có:
Điều kiện để căn thức có nghĩa làhay
Do đó:
Nếu b>0 thì
Nếuthì
- Điều kiện để
có nghĩa là
hay
Cách 1.
=
Cách 2. Biến mẫu thành một bình phương rồi áp dụng quy tắc khai phương một thương: - Điều kiện để
có nghĩa là
hay xy>0.
Do đó
a. 2√5