Tìm x
3^x .2^2 + 3^x . 5 = 81
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a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
a, xem lại đề , sửa rồi thì báo cho tui
b, \(\left(x+\frac{2}{5}\right)^5=\left(x+\frac{2}{5}\right)^3\)
\(\Rightarrow\left(x+\frac{2}{5}\right)^5-\left(x+\frac{2}{5}\right)^3=0\)
\(\Rightarrow\left(x+\frac{2}{5}\right)^3.\left[\left(x+\frac{2}{5}\right)^2-1\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+\frac{2}{5}\right)^3=0\\\left(x+\frac{2}{5}\right)^2-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{2}{5}\\\left(x+\frac{2}{5}\right)^2=1\end{cases}}}\)
Ta có \(\left(x+\frac{2}{5}\right)^2=1\)
\(\Rightarrow\hept{\begin{cases}x+\frac{2}{5}=1\\x+\frac{2}{5}=-1\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{5}\\x=-\frac{7}{5}\end{cases}}}\)
Vậy \(x\in\text{{}-\frac{2}{5};\frac{3}{5};-\frac{7}{5} \)}
a, \(x=\dfrac{3}{4}-1=\dfrac{3}{4}-\dfrac{4}{4}=-\dfrac{1}{4}\)
b, \(x=\dfrac{1}{5}-4=\dfrac{1}{5}-\dfrac{20}{5}=-\dfrac{19}{5}\)
c, \(x=2+\dfrac{1}{5}=\dfrac{10}{5}+\dfrac{1}{5}=\dfrac{11}{5}\)
d, \(x=\dfrac{1}{81}-\dfrac{5}{3}=\dfrac{1}{81}-\dfrac{135}{81}=-\dfrac{134}{81}\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
a: \(\Leftrightarrow2^3< 2^x< 2^4\)
=>3<x<4
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
=>12-x=4
hay x=8
c: \(\Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{2}{5}\right)^2=\left(\dfrac{2}{5}\right)^5\)
=>x>5
d: \(\Leftrightarrow3x-1=-4\)
=>3x=-3
hay x=-1
a: \(\Leftrightarrow2^3< 2^x< 2^4\)
=>3<x<4
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
=>12-x=4
hay x=8
c: \(\Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{2}{5}\right)^2=\left(\dfrac{2}{5}\right)^5\)
=>x>5
d: \(\Leftrightarrow3x-1=-4\)
=>3x=-3
hay x=-1
3^x.2^2+3^x.5=81
3^x.4+3^x.5=81
3^x.(4+5)=81
3^x.9=81
3^x=81:9
3^x=9
3^x=3^3
suy ra x=3