Khử mẫu biểu thức chứa căn ms đúng

\(\sqrt{\frac{\left(1+\sqrt{2}\right)^3}{27}}=\sqrt{\frac{\left(1+\sqrt{2}\right)^2\cdot\left(1+\sqrt{2}\right)}{3^2\cdot3}}=\frac{1+\sqrt{2}}{3}\cdot\sqrt{\frac{1+\sqrt{2}}{3}}\)

\(=\frac{1+\sqrt{2}}{3}\cdot\frac{\sqrt{3\cdot\left(1+\sqrt{2}\right)}}{3}=\frac{1+\sqrt{2}}{9}\cdot\sqrt{3+3\sqrt{2}}\)

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

Lời giải:

\(\sqrt{\frac{(1+\sqrt{2})^3}{27}}=\sqrt{\frac{(1+\sqrt{2})^3}{3^3}}=\sqrt{\frac{3(1+\sqrt{2})^3}{3^4}}\)

\(=\frac{(1+\sqrt{2})\sqrt{3+3\sqrt{2}}}{9}\)

\(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{(ab)^2(\frac{1}{a}+\frac{1}{b})}=\sqrt{ab^2+a^2b}\)

Lời giải:

Ta có:

\(\frac{\sqrt{3}+\sqrt{5}}{(\sqrt{5}+1)(\sqrt{3}-1)}=\frac{(\sqrt{3}+\sqrt{5})(\sqrt{5}-1)(\sqrt{3}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)(\sqrt{3}-1)(\sqrt{3}+1)}\)

\(=\frac{(\sqrt{3}+\sqrt{5})(\sqrt{5}-1)(\sqrt{3}+1)}{(5-1)(3-1)}=\frac{(\sqrt{3}+\sqrt{5})(\sqrt{5}-1)(\sqrt{3}+1)}{8}\)

;

a) \(\frac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right).\left(1-\sqrt{2}+\sqrt{3}\right)}.\)

\(=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(\sqrt{2}-\sqrt{3}\right)^2}=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(5-2\sqrt{6}\right)}\)

\(=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{-4+2\sqrt{6}}=\frac{1-\sqrt{2}+\sqrt{3}}{-2\sqrt{2}+2\sqrt{3}}\)

\(=\frac{\left(1-\sqrt{2}+\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}{-2\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=\frac{\left(1-\sqrt{2}+\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}{-2.\left(2-3\right)}\)\(=\frac{\left(1-\sqrt{2}+\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}{2}\)

Căn thức ở mẫu đã được trục rồi.

Nếu cần thì phá ngoặc phần tử số ra.

b) Nhân cả tử số và mẫu số cho \(\sqrt{a+3}-\sqrt{a-3}\)thì mẫu số có giá trị là (a + 3) - (a - 3) = 6; tử số có giá trị là \(\left(\sqrt{a+3}-\sqrt{a-3}\right)^2\). Khi đó, căn thức ở mẫu đã được trục đi rồi. Sau đó bạn phá ngoặc phần tử số ra.

EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.

1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)

2) \(\sqrt{\frac{a-4}{2\left(\sqrt{a}-2\right)}}=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}}\)

\(=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(a-4\right)}}\)

3) \(\sqrt{\frac{1}{a\left(1-\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-3\right)}}=\sqrt{-\frac{1+\sqrt{3}}{2a}}\)

4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)