(2x + 3/5)2 - 9/25 = 0
(2x +3/5)2= 9/25
2x+3/5=3/5
2x=0
=) x=0
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
\(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\frac{3}{5}\right)^{^2}-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
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
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
tim x
tìm x\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
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) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(100-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{27}}+\left(1,2.0,5\right):\frac{3}{5}\)
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} 3left(x-frac{1}{2}right)-5left(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}xfrac{5}{12}Help me, please
Đọc tiếp
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