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câu e mình viết sai đề, mk sửa lại nhé , với mình bổ sung câu f
e) \(\dfrac{2}{\sqrt[3]{4}+\sqrt[3]{5}}\)
f) \(\dfrac{1}{2-\dfrac{\sqrt[3]{3}}{2}}\)
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=1/(√3+√2+1)=(√3-(√2+1)/[3-(√2+1)^2]
=(√3-√2-1)/(3-(3+2√2)
=(√3-√2-1)/(-2√2)
=-(√6-2-√2)/4
=(2+√2-√6)/4
a: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{7+2\sqrt{10}-3}=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{4+2\sqrt{10}}\)
\(=\dfrac{-\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(4-2\sqrt{10}\right)}{24}\)
b: \(=\dfrac{2+\sqrt{3}+\sqrt{5}}{4-8+2\sqrt{15}}=\dfrac{2+\sqrt{3}+\sqrt{5}}{2\sqrt{15}-4}\)
\(=\dfrac{\left(2+\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+4\right)}{44}\)
a. \(\dfrac{1}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}\right)^2-2}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{5+3-2-2\sqrt{15}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{6-2\sqrt{15}}=\dfrac{\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{15}\right)}{\left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right)2}=\dfrac{3\sqrt{5}-3\sqrt{3}-3\sqrt{2}+5\sqrt{3}-3\sqrt{5}-\sqrt{30}}{\left(9-15\right).2}=\dfrac{2\sqrt{3}-3\sqrt{2}-\sqrt{30}}{-12}\)b. \(\dfrac{1}{2-\sqrt{3}-\sqrt{5}}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}-\sqrt{5}\right)\left(2-\sqrt{3}+\sqrt{5}\right)}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}\right)^2-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{4-4\sqrt{3}+3-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{2-4\sqrt{3}}=\dfrac{\left(2-\sqrt{3}+\sqrt{5}\right)\left(1+2\sqrt{3}\right)}{2\left(1-2\sqrt{3}\right)\left(1+2\sqrt{3}\right)}=\dfrac{2+4\sqrt{3}-\sqrt{3}-6+\sqrt{5}+2\sqrt{15}}{2.\left(1-12\right)}=\dfrac{3\sqrt{3}+\sqrt{5}+2\sqrt{15}-4}{-22}\)
a) \(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}\dfrac{\sqrt{2}+2+\sqrt{6}}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{\sqrt{2}+2+\sqrt{6}}{2\sqrt{2}+3-3}=\dfrac{\sqrt{2}+2+\sqrt{6}}{2\sqrt{2}}=\dfrac{1+\sqrt{2}+\sqrt{3}}{2}\)
b) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}+5-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}}=\dfrac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{2\sqrt{6}\cdot\sqrt{6}}=\dfrac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12}\)
bài 2:
a: \(\dfrac{25}{5-2\sqrt{3}}=\dfrac{125+10\sqrt{3}}{13}\)
b: \(\dfrac{8}{\sqrt{5}+2}=8\sqrt{5}-32\)
c: \(\dfrac{6}{2\sqrt{3}-\sqrt{7}}=\dfrac{12\sqrt{3}+6\sqrt{7}}{5}\)
d: \(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\dfrac{\sqrt{6}}{2}\)
Lời giải:
a) \(\frac{1}{1-\sqrt[3]{5}}=\frac{1+\sqrt[3]{5}+\sqrt[3]{5^2}}{(1-\sqrt[3]{5})(1+\sqrt[3]{5}+\sqrt[3]{25})}\) \(=\frac{1+\sqrt[3]{5}+\sqrt[3]{25}}{1^3-5}=\frac{1+\sqrt[3]{5}+\sqrt[3]{25}}{-4}\)
b)
\(\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}}=\frac{\sqrt[3]{2^2}-\sqrt[3]{6}+\sqrt[3]{3^2}}{(\sqrt[3]{2}+\sqrt[3]{3})(\sqrt[3]{2^2}-\sqrt[3]{6}+\sqrt[3]{3^2})}\) \(=\frac{\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}}{2+3}=\frac{\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}}{5}\)
c)
\(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{2^2}+\sqrt[3]{2}+1)}=\frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}-\sqrt{2}-1\right)}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-1}{3-\left(\sqrt{2}+1\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}-1}{-2\sqrt{2}}=\dfrac{\left(\sqrt{3}-\sqrt{2}-1\right)\sqrt{2}}{-2\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}-2-\sqrt{2}}{-4}\)
\(=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)