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Theo đề ra ,ta có :
- 1 / 12 < x < 1 / 8 mà x có giá trị nguyên
=> x = 0
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)
a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=9\)
b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)
\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)
\(\Leftrightarrow x=66\)
\(\left(\frac{2}{3}\right)^{x-2}=\left(\frac{16}{81}\right)^{x+1}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{x-2}=\left[\left(\frac{2}{3}\right)^4\right]^{x+1}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{x-2}=\left(\frac{2}{3}\right)^{4\left(x+1\right)}\)
\(\Leftrightarrow x-2=4x+4\)
\(\Leftrightarrow-3x=6\Leftrightarrow x=-2\)
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+...+\frac{1}{16}.\frac{16.\left(16+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=\frac{\left(17-2+1\right).\left(17+2\right)}{2}:2\)
\(=76\)
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}\left[\frac{\left(2+1\right)2}{2}\right]+\frac{1}{3}\left[\frac{\left(3+1\right)3}{3}\right]+...+\frac{1}{16}\left[\frac{\left(16+1\right)16}{2}\right]\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2+2+1+3+1+...+16+1}{2}\)
\(=\frac{\left(1+1+1+..15cs.+1\right)+\left(2+3+...+16\right)+2}{2}\)
\(=\frac{15+135+2}{2}\)
\(=\frac{152}{2}\)\(=76\)
\(\left(\frac{2}{3}\right)^x=\frac{16}{81}\)
\(\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^4\)
\(\Rightarrow x=4\)
\(\left(x+\frac{1}{2}\right)^4=\frac{16}{81}\)
\(\Rightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\\\left(x+\frac{1}{2}\right)^4=\left(\frac{-2}{3}\right)^4\end{cases}}\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{3}\\x+\frac{1}{2}=\frac{-2}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-7}{6}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{6};\frac{-7}{6}\right\}\)
_Chúc bạn học tốt_