Các câu hỏi tương tự
\((x+\frac{1}{2})\times(\frac{2}{3}-2x)=0\)
\(\left\{x+\frac{1}{5}\right\}^2+\frac{17}{25}=\frac{26}{25}\)
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)0
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
tim x
tìm x\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(C=\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)= ?
\(C=\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)= ?