Tìm x \(\in\) Za,\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right) x \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)b, \(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}...
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Tìm x \(\in\) Z
a,\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
b, \(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{3}\right)\)
c, \(x:\left(\frac{1}{2}\right)^2=\frac{-1}{2}\)
d,\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)
e,\(\frac{3^2.3^8}{27^3}=3^x\)
f,\(2^{x-1}=\left(16\right)^5\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21