x=\(\dfrac{52}{63}\) x -\(\dfrac{9}{26}\)
x=-1\(\dfrac{5}{7}\) =-\(\dfrac{2}{7}\)
x=\(\dfrac{52}{63}\) x -\(\dfrac{9}{26}\)
x=-1\(\dfrac{5}{7}\) =-\(\dfrac{2}{7}\)
\(\text{Tìm x, biết:}\)
\(a\)) \(20\text{%}x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(b\)) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(c\)) \(\left(x-\dfrac{3}{4}\right)\left(4+3x\right)=0\)
\(d\)) \(x-\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{-26}{5}\)
\(e\)) \(50\text{%}x+\dfrac{2}{3}x=x-5\)
\(g\)) \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}.\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
Tìm x liên quan đến lũy thừa:
1, \(\left(3x-\dfrac{1}{5}\right)^2=\left(\dfrac{-3}{25}\right)^2\)
2, \(\left(2x-\dfrac{1}{3}\right)^2=\left(\dfrac{-2}{9}\right)^2\)
3, \(\left(\dfrac{1}{3}-x\right)^2=\dfrac{9}{25}\)
4, \(\left(5-x\right)^2=25\)
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
Tìm số nguyên x, biết: ( \(\dfrac{31}{20}-\dfrac{26}{45}\) ) . \(\dfrac{-36}{35}< x< \left(\dfrac{51}{56}+\dfrac{8}{21}+\dfrac{1}{3}\right).\dfrac{8}{13}\)
Tìm x:
\(\dfrac{3}{8}\left(x-\dfrac{2}{3}\right)-\dfrac{2}{9}.\left(3x-\dfrac{1}{2}\right)=-\dfrac{7}{36}\)
tìm x biết
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}.x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
|x|\(-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|2.x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
giúp mk vs nhanh lên mình đang bận
Tìm x, biết:
\(a,\dfrac{1}{3}:\left(2x-1\right)=\dfrac{-1}{6}\)
\(b,\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(c,\dfrac{x}{8}=\dfrac{9}{4}\)
\(d,\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(e,4,5x-6,2x=6,12\)
\(h,11,4-\left(x-3,4\right)=-16,2\)
Cho mk hỏi :
tìm x,biết
a,\(-3\dfrac{1}{2}\)x -0,75-1,25x=\(\left(\dfrac{-1}{2}\right)^2:\dfrac{-3}{4}+\dfrac{1}{6}\)
b, \(\dfrac{-2}{3}-\left(\dfrac{x}{2}-75\%\right)=\left(\dfrac{3}{-4}-\dfrac{9}{8}\right)^2:\dfrac{-3}{32}-1\dfrac{1}{3}\)
Tìm x, biết
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)