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1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
Chúc bạn học tốt!
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a) \(\left|x\right|-15=6\Rightarrow\left|x\right|=21\Rightarrow\left[{}\begin{matrix}x=21\\x=-21\end{matrix}\right.\)
b) \(\left|x\right|+4=0\Rightarrow\left|x\right|=-4\Rightarrow x\in\varnothing\)
c) \(x^2-16=0\Rightarrow x^2=16=4^2\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
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a) \(\frac{x}{6}=\frac{8}{3}\Rightarrow x=\frac{8\cdot6}{3}=16\)
b) \(\frac{-3}{x}=\frac{15}{7}\Rightarrow x=-\frac{3\cdot7}{15}=-\frac{7}{5}\)
c) \(\frac{0,1}{5}=\frac{x}{15}\Rightarrow x=\frac{15\cdot0,1}{5}=0,3\)
d) \(\frac{x-3}{3}=\frac{4}{5}\Leftrightarrow x-3=\frac{4\cdot3}{5}=\frac{12}{5}\)
\(\Rightarrow x=\frac{12}{5}+3=\frac{27}{5}\)
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\(x:\dfrac{13}{16}=\dfrac{5}{-8}\\ x=\dfrac{5}{-8}.\dfrac{13}{16}=\dfrac{65}{-128}\)
\(x.\dfrac{-1}{2}=\dfrac{-4}{5}\\ x=\dfrac{-4}{5}:\dfrac{-1}{2}=\dfrac{8}{5}\)
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a,3(x-2)+4(x-3)-6(x+8)=15
3x-6+4x-12-6x+48-15=0
x-81=0
=>x=81
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a: \(=\dfrac{2^{30}\cdot3^{18}}{3^{16}\cdot2^{31}}=3^2\cdot\dfrac{1}{2}=\dfrac{9}{2}\)
đề bại bạn ơi?
đề bài là gì v b