⇒x=1/23+1/27 x=1/8+1/27 ⇒x=35/216
⇒x=1/23+1/27 ⇒x=1/8+1/27
x=35/216
\(x-\left(\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(x-\dfrac{1}{8}\) \(=\dfrac{1}{27}\)
\(x\) \(=\dfrac{1}{27}+\dfrac{1}{8}=\dfrac{35}{216}\)
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\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
Tìm x:a) dfrac{2}{3}+left(x-dfrac{1}{2}right)^3dfrac{19}{27}b) left(dfrac{3}{2}right)^{2x-1}:left(dfrac{27}{8}right)^3dfrac{81}{16}c) dfrac{1}{2}.2^x+4.2^x9.2^5d) text{12 - (2x +1)}^2-69
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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})^3:3dfrac{-1}{81}c,(dfrac{2}{3}-x)^21:dfrac{4}{9}d,(2x-dfrac{1}{5})^2+dfrac{16}{25}1e,(dfrac{2}{5}-3x)^2-dfrac{1}{5}dfrac{4}{25}
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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 xa) |x-2|-dfrac{3}{5}dfrac{1}{2}b) (x-dfrac{7}{3}):dfrac{-1}{3}0,4c) |x-3|5d) (2x+3)^225e) dfrac{3}{4}+dfrac{1}{4}: xdfrac{2}{5}f) (x-dfrac{1}{2})^3dfrac{1}{27}Làm nhanh giúp mik vs ạ(đang rất cần, cảm ơn nhiều ạaa!!!!!)
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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!!!!!)
\(\dfrac{1}{3}\)x(2x-\(\dfrac{1}{3}\))\(^2\)-\(\dfrac{25}{27}\)=0
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}) + 1dfrac{2}{3}] - dfrac{-5}{27}
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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\(\dfrac{2}{3}\)] - \(\dfrac{-5}{27}\)
\(x^2-19=5.9;\left(2x+1\right)^3=-0,001;\left(\dfrac{5}{6}\right)^{2x-1}=\left(\dfrac{5}{6}\right)^5;\left(\dfrac{1}{3}x-\dfrac{2}{3}\right)^3=27;\left(\dfrac{1}{32}\right)^x=\left(\dfrac{1}{2}\right)^{15}\)
MH
20 tháng 7 2021 lúc 21:10
(x - \(\dfrac{2}{3}\))3 = \(\dfrac{-1}{27}\)