Violympic toán 9
Các câu hỏi tương tự
{\(\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\)
1\(\frac{1}{x+y}-\frac{1}{x-y}=\frac{3}{8}\)
\(\left(\frac{b}{b+8}-\frac{4b}{\left(\sqrt[3]{b}+2\right)^3}\right).\left(\frac{1+2\sqrt[3]{\frac{1}{b}}}{1-2\sqrt[3]{\frac{1}{b}}}\right)-\frac{24}{b+8}\)
tìm x, biết :
a, \(1-(5\frac{3}{8}+x-7\frac{5}{24}):(-16\frac{2}{3})=0\)
b, \((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
c, \(1\frac{3}{5}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}.x=\frac{16}{5}\)
\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{1}{y+2}=2\\\frac{8}{x}-\frac{3}{y+2}=1\end{matrix}\right.\)
a, \(\frac{3}{5}.x-\frac{1}{2}=\frac{1}{7}\)
b, \(\frac{1}{4}+\frac{1}{3}:3x=-5\)
c, \(\frac{1}{3}.x+\frac{2}{5}\left(x+1\right)=0\)
d, \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
1,Rút gọn:
a, \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+2}\)
b,\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\frac{8}{x-3}+\frac{1}{2\left|y\right|-3}=5\\\frac{4}{x-3}+\frac{1}{2\left|y\right|-3}=3\end{matrix}\right.\)
Cho 4(a+b+c)=3abc
CMR \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge\frac{3}{8}\)
13 tháng 8 2020 lúc 20:35
Giải pt
\(\frac{1}{x+1}+\frac{3}{2x+1}=\frac{8}{x-2}\)