`1/(-27)=(1/(-x))^3`
`(1/(-3))^3=(1/(-x))^3`
`-3=-x`
`x=3`
Vậy `x=3`
\(\left(\dfrac{1}{-x}\right)^3=\dfrac{1}{-27}=\left(\dfrac{1}{-3}\right)^3\Rightarrow-x=-3\Leftrightarrow x=3\)
`1/(-27)=(1/(-x))^3`
`(1/(-3))^3=(1/(-x))^3`
`-3=-x`
`x=3`
Vậy `x=3`
\(\left(\dfrac{1}{-x}\right)^3=\dfrac{1}{-27}=\left(\dfrac{1}{-3}\right)^3\Rightarrow-x=-3\Leftrightarrow x=3\)
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
\(\left(\dfrac{-2}{3}\right)^2.x=\left(\dfrac{-2}{3}\right)^5\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)
\(\dfrac{4}{9}:x=3\dfrac{1}{3}:2,25\)
\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\)
a,2.(\(\dfrac{1}{4}\)+x)\(^3\)=(\(-\dfrac{27}{4}\))
b,(x+\(\dfrac{1}{2}\))\(^3\):3=\(\dfrac{-1}{81}\)
c,(\(\dfrac{2}{3}\)-x)\(^2\)=1:\(\dfrac{4}{9}\)
d,(2x-\(\dfrac{1}{5}\))\(^2\)+\(\dfrac{16}{25}\)=1
e,(\(\dfrac{2}{5}\)-3x)\(^2\)-\(\dfrac{1}{5}\)=\(\dfrac{4}{25}\)
Bài 2:Tìm x
a) |x-2|-\(\dfrac{3}{5}\)=\(\dfrac{1}{2}\)
b) (x-\(\dfrac{7}{3}\)):\(\dfrac{-1}{3}\)=0,4
c) |x-3|=5
d) (2x+3)\(^2\)=25
e) \(\dfrac{3}{4}\)+\(\dfrac{1}{4}\): x=\(\dfrac{2}{5}\)
f) (x-\(\dfrac{1}{2}\))\(^3\)=\(\dfrac{1}{27}\)
Làm nhanh giúp mik vs ạ(đang rất cần, cảm ơn nhiều ạaa!!!!!)
tìm x
a) x: 27=3,6
b)\(\dfrac{2x+1}{-27}\) =\(\dfrac{-3}{2x+1}\)
\(\dfrac{1}{3}\)x(2x-\(\dfrac{1}{3}\))\(^2\)-\(\dfrac{25}{27}\)=0
\(\dfrac{x-1}{3}=\dfrac{27}{x-1}\)
a \(\dfrac{-4}{7}\) - \(\dfrac{5}{13}\) x \(\dfrac{-39}{25}\) + \(\dfrac{-1}{42}\) : \(\dfrac{-5}{6}\)
b \(\dfrac{2}{9}\) x [\(\dfrac{4}{45}\): ( \(\dfrac{1}{5}\) - \(\dfrac{2}{15}\)) + 1\(\dfrac{2}{3}\)] - \(\dfrac{-5}{27}\)
\(x^2-19=5.9;\left(2x+1\right)^3=-0,001;\left(\dfrac{5}{6}\right)^{2x-1}=\left(\dfrac{5}{6}\right)^5;\left(\dfrac{1}{3}x-\dfrac{2}{3}\right)^3=27;\left(\dfrac{1}{32}\right)^x=\left(\dfrac{1}{2}\right)^{15}\)