Question

[Phương Uyên]

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,\(2^{x-1}=\left(16\right)^5\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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