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a)5x+5x+2=650
\(\Rightarrow5^x\left(1+5^2\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
b)\(3^{x-1}+5\cdot3^{x-1}=162\)
\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}\cdot6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
mk giải ở dưới được 2 câu rùi nhưng ko chắc cau 2 ~~~
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a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)
\(\frac{X}{5}=\frac{1}{5}\)
Vậy \(X=1\)
b)\(X-\frac{41}{5}=\frac{-2}{3}\)
\(X=\frac{-2}{3}+\frac{41}{5}\)
\(X=\frac{113}{15}\)
Vậy \(X=\frac{113}{15}\)
c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)
\(\frac{31}{5}-X=\frac{131}{30}\)
\(X=\frac{31}{5}-\frac{131}{30}\)
\(X=\frac{11}{6}\)
Vậy \(X=\frac{11}{6}\)
d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)
\(\frac{9}{X}=\frac{1}{20}\)
\(X=9:\frac{1}{20}\)
\(X=180\)
Vậy \(X=180\)
Hc tốt
\(a.x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=0\\x_2=2\end{matrix}\right.\)
\(b.x^{50}=x^2\Leftrightarrow x^{50}-x^2=0\Leftrightarrow x^2\left(x^{48}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^{48}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{48}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=0\\x_2=1\\x_3=-1\end{matrix}\right.\)
\(c.\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\Leftrightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)
\(\Leftrightarrow\left(3x-1\right)^{10}\left[1-\left(3x-1\right)^{10}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3x-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1}{3}\\x_2=\dfrac{2}{3}\end{matrix}\right.\)