\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(\Rightarrow x=0\)
\(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\frac{3}{5}\right)^{^2}-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
Tìm x
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
giải nhanh hộ mình với, mai mình nộp rồi
tìm x\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
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) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
Tìm x, biết:
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(3(3x-\frac{1}{2})^3+\frac{1}{9}=0\) \(\frac{3}{4}-2.|2x-\frac{2}{3}|=2\) \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
Help me, please
\((x+\frac{1}{2})\times(\frac{2}{3}-2x)=0\)
\(\left\{x+\frac{1}{5}\right\}^2+\frac{17}{25}=\frac{26}{25}\)
