\(A=\frac{\sqrt{18}}{\sqrt{8}}+\frac{\sqrt{8}}{\sqrt{50}}=\frac{\sqrt{18}.\sqrt{50}+\sqrt{8}.\sqrt{8}}{\sqrt{8}.\sqrt{50}}=\frac{\sqrt{900}+\sqrt{64}}{\sqrt{400}}=\frac{30+8}{20}=\frac{38}{20}=\frac{19}{10}\)
\(M=\frac{3\sqrt{18}+2\sqrt{50}-3\sqrt{72}+4\sqrt{98}}{2\sqrt{8}+3\sqrt{12}-5\sqrt{20}}\)
Các bạn giải các bước ra luôn nha, thanks
(\(18\frac{1}{3}:\sqrt{225}+8\frac{2}{3}.\sqrt{\frac{49}{2^2}}\)): [(\(12\frac{1}{3}+8\frac{6}{7}\)) - \(\frac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\)] : \(\frac{1704}{445}\)
\(B=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^{^2}}}}\)RÚT GỌN B với a>8
Tính:
a) A=\(\sqrt{18}+\sqrt{50}-\frac{1}{2}\sqrt{98}\)
b) B=\(\left(2\sqrt{3}+7\right)\left(2\sqrt{3}-7\right)\)
c) C=\(\sqrt{7-2\sqrt{10}+\sqrt{2}}\)
Trong các số sau đây sô nào bằng \(\frac{3}{5}\)
a,\(\sqrt{\frac{3^2}{5^2}}\)
b,\(\frac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
c,\(\frac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}\)
bài 1 tính
a)\(\frac{-8}{18}-\frac{15}{27}\)
b. \(\frac{1}{2}.\sqrt{100-\sqrt{\frac{1}{16}+\left(\frac{1}{3}\right)^0}}\)
c. \(\frac{5^4.20}{25^5.4^5}^4\)
1. Chứng minh:\(\left(\frac{x\sqrt{x}+27y\sqrt{y}}{3\sqrt{x}+9\sqrt{y}}-\sqrt{xy}\right).\left(\frac{3\sqrt{x}+9\sqrt{y}}{9y-x}\right)^2>\sqrt{8}\)
2. Rút gọn A= \(\frac{\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}}{\sqrt{a+\sqrt{2a-1}}-\sqrt{a-\sqrt{2a-1}}}\)
Chứng minh rằng:
a)\(\sqrt{1}+\sqrt{2}+...+\sqrt{8}< 24\)
b)\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>10\)
c)\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}< 30\)
\(\frac{4.\sqrt{65465}:\sqrt[33]{56}}{9.\sqrt{65475}:\sqrt[8]{5}}\)
Tính
