Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)
\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)
\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)
\(=\frac{2\sqrt{3}}{3}+2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)
\(=2\sqrt{3}\)
1/\(\sqrt{8-2\sqrt{15}}-\sqrt{21-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
Bạn tự làm tiếp
2/ \(\frac{4}{\sqrt{7-4\sqrt{3}}}-\frac{4}{7-4\sqrt{3}}=\frac{4}{\sqrt{\left(2-\sqrt{3}\right)^2}}-\frac{4}{\left(2-\sqrt{3}\right)^2}=\frac{4}{2-\sqrt{3}}-\frac{4}{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{8-4\sqrt{3}-4}{\left(2-\sqrt{3}\right)^2}=\frac{4-4\sqrt{3}}{\left(2-\sqrt{3}\right)^2}\) đến đây ko rút gọn được nữa, nghi bạn chép sai đề.
Tử số của phân số thứ hai là 4 hay 1 vậy?
3/ \(\frac{\sqrt{8+2\sqrt{15}}-\sqrt{4-2\sqrt{3}}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{3+\sqrt{5}}{2}\)
4/ \(\frac{10}{\sqrt{\left(\sqrt{5}-2\right)^2}}-\frac{12}{\sqrt{\left(3+\sqrt{5}\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{3+\sqrt{5}}+\frac{20}{\sqrt{5}-1}\)
\(=\frac{10\left(\sqrt{5}+2\right)}{1}-\frac{12\left(3-\sqrt{5}\right)}{4}+\frac{20\left(\sqrt{5}+1\right)}{4}=16+18\sqrt{5}\)
\(\frac{10}{\sqrt{5}-2.\sqrt{5}.2+4}-\frac{12}{\sqrt{\sqrt{5}+2.\sqrt{5}.3+9}}+\frac{20}{\sqrt{5-2.\sqrt{5}.1+1}}=\frac{10}{\left(\sqrt{5}-2\right)^2}-\frac{12}{\sqrt{\left(\sqrt{5}+3\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{\sqrt{5}+3}+\frac{20}{\sqrt{5}-1}=\frac{10\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right).\left(\sqrt{5}+2\right)}-\frac{12.\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right).\sqrt{5}-3\left(\right)}+\frac{20.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\frac{10\sqrt{5}-20}{5-4}-\frac{12\sqrt{5}-36}{5-9}+\frac{20\sqrt{5}+20}{5-1}\\=\frac{40\sqrt{5}-80+12\sqrt{5}+36+20\sqrt{5}+20}{4}=\\ 18\sqrt{5}-6\)
a) Ta có: \(A=\frac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\cdot\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{3}+\sqrt{5}\)
b) Ta có: \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}+\frac{\sqrt{5}-\sqrt{4}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\sqrt{5}-2+\sqrt{6}-\sqrt{5}\)
\(=-1+\sqrt{6}\)
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
