\(\left(\dfrac{6}{5}\right)^x=\left(\dfrac{6}{5}\right)^3\Leftrightarrow x=3\)
Vậy \(x=3\)
\(\left(\dfrac{6}{5}\right)^x=\left(\dfrac{6}{5}\right)^3\Leftrightarrow x=3\)
Vậy \(x=3\)
Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{1}{2}\)x = \(\left(-\dfrac{7}{12}\right)\) . \(1\dfrac{2}{5}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2\) = \(\dfrac{9}{4}\)
c) (1,25 - \(\dfrac{4}{5}\)x)3 = -125
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.......+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)(x ϵ N*)
Tìm x :
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
\(\dfrac{5^{x+1}}{125}\)=\(\dfrac{1}{25^{x-2}}\)
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
Tìm x
Tìm x biết:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\left(x\inℕ^∗\right)\)
Giúp mik với, giải chi tiết dùm nhe.
Tìm x,y ∈ \(Z\) , biết :
a) \(\dfrac{x}{5}+1=\dfrac{x}{y-1}\)
b) \(\dfrac{2}{x}+\dfrac{y}{3}=\dfrac{1}{6}\)
c) \(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
Tìm x :
a, \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) +x) = \(\dfrac{2}{3}\) b, x+\(\dfrac{2}{3}\) =\(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\) c, x-[\(\dfrac{17}{2}\) -(\(\dfrac{-3}{7}\) + \(\dfrac{5}{3}\) )]=\(\dfrac{-1}{3}\)
d,\(\dfrac{9}{2}\) - [\(\dfrac{2}{3}\) - (x+\(\dfrac{7}{4}\) )] =\(\dfrac{-5}{4}\)
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