Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}\dfrac{\sqrt{2}+2+\sqrt{6}}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{\sqrt{2}+2+\sqrt{6}}{2\sqrt{2}+3-3}=\dfrac{\sqrt{2}+2+\sqrt{6}}{2\sqrt{2}}=\dfrac{1+\sqrt{2}+\sqrt{3}}{2}\)
b) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}+5-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}}=\dfrac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{2\sqrt{6}\cdot\sqrt{6}}=\dfrac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12}\)
a) \(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}+1-\sqrt{2}}{\left(\sqrt{3}+1+\sqrt{2}\right)\left(\sqrt{3}+1-\sqrt{2}\right)}\)
= \(\dfrac{\sqrt{3}+1-\sqrt{2}}{\left(\sqrt{3}+1\right)^2-2}=\dfrac{\left(\sqrt{3}+1-\sqrt{2}\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
= \(\dfrac{3-\sqrt{3}+\sqrt{3}-1-\sqrt{6}+\sqrt{2}}{2\left(3-1\right)}\) = \(\dfrac{2-\sqrt{6}+\sqrt{2}}{4}\)
b) \(\dfrac{1}{\sqrt{5}+2-\sqrt{3}}=\dfrac{\sqrt{5}+2+\sqrt{3}}{\left(\sqrt{5}+2\right)^2-3}\) = \(\dfrac{\sqrt{5}+\sqrt{3}+2}{4\sqrt{5}+6}\)
= \(\dfrac{\left(\sqrt{5}+\sqrt{3}+2\right)\left(4\sqrt{5}-6\right)}{\left(4\sqrt{5}+6\right)\left(4\sqrt{5}-6\right)}\) = \(\dfrac{20-6\sqrt{5}+4\sqrt{15}-6\sqrt{3}+8\sqrt{5}-12}{\left(4\sqrt{5}\right)^2-36}\)
= \(\dfrac{8+2\sqrt{5}-6\sqrt{3}+4\sqrt{15}}{44}\) = \(\dfrac{2\left(4+\sqrt{5}-3\sqrt{3}+2\sqrt{15}\right)}{2\left(22\right)}\)
= \(\dfrac{4+\sqrt{5}-3\sqrt{3}+2\sqrt{15}}{22}\)
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)
\(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{2^2}+\sqrt[3]{2}+1)}=\frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1\)
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
Nhat Linh bị nhầm câu cuối:
\(\dfrac{y+b\sqrt{y}}{b.\sqrt{y}}=\dfrac{y\sqrt{y}+b.y}{b.y}=\dfrac{\sqrt{y}+b}{b}.\)
c) \(\dfrac{3\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\dfrac{3\sqrt{3}}{\left(\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}\right)}=\dfrac{3\sqrt{3}\left(\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}\right)}{\left(\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}\right)\left(\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}\right)}\) = \(\dfrac{3\sqrt{6}+9-3\sqrt{15}}{\left(\sqrt{2}+\sqrt{3}\right)^2-5}\) = \(\dfrac{3\sqrt{6}+9-3\sqrt{15}}{2+2\sqrt{6}+3-5}=\dfrac{3\sqrt{6}+9-3\sqrt{15}}{2\sqrt{6}}\)
= \(\dfrac{\left(3\sqrt{6}+9-3\sqrt{15}\right)\sqrt{6}}{2\sqrt{6}.\sqrt{6}}\) = \(\dfrac{18+9\sqrt{6}-9\sqrt{10}}{12}\)
= \(\dfrac{3\left(6+3\sqrt{6}-3\sqrt{10}\right)}{3.4}=\dfrac{6+3\sqrt{6}-3\sqrt{10}}{4}\)
d) \(\dfrac{4}{1+\sqrt{2}+\sqrt{3}}=\dfrac{4}{\left(\left(1+\sqrt{2}\right)+\sqrt{3}\right)}=\dfrac{4\left(\left(1+\sqrt{2}\right)-\sqrt{3}\right)}{\left(\left(1+\sqrt{2}\right)+\sqrt{3}\right)\left(\left(1+\sqrt{2}\right)-\sqrt{3}\right)}\)
= \(\dfrac{4+4\sqrt{2}-4\sqrt{3}}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{4+4\sqrt{2}-4\sqrt{3}}{1+2\sqrt{2}+1-3}\) = \(\dfrac{4+4\sqrt{2}-4\sqrt{3}}{2\sqrt{2}}\)
\(\dfrac{\left(4+4\sqrt{2}-4\sqrt{3}\right)\sqrt{2}}{2\sqrt{2}\sqrt{2}}=\dfrac{4\sqrt{2}+8-4\sqrt{6}}{4}\) = \(\dfrac{4\left(\sqrt{2}+4-\sqrt{6}\right)}{4}=\sqrt{2}+4-\sqrt{6}\)
câu a thôi nha
câu b:\(\dfrac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}{12}=\dfrac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\)
câu c,d tương tự câu b thôi
bản chất lười =))
\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
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. \(\dfrac{1}{2-\sqrt{3}-\sqrt{5}}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}-\sqrt{5}\right)\left(2-\sqrt{3}+\sqrt{5}\right)}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}\right)^2-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{4-4\sqrt{3}+3-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{2-4\sqrt{3}}=\dfrac{\left(2-\sqrt{3}+\sqrt{5}\right)\left(1+2\sqrt{3}\right)}{2\left(1-2\sqrt{3}\right)\left(1+2\sqrt{3}\right)}=\dfrac{2+4\sqrt{3}-\sqrt{3}-6+\sqrt{5}+2\sqrt{15}}{2.\left(1-12\right)}=\dfrac{3\sqrt{3}+\sqrt{5}+2\sqrt{15}-4}{-22}\)