Question

[Roxie]

1)tim x bit a)(1/2-1/3).6^x+6^x+2=6^15+6^18

b)3^x+3^x+2=2430

c)(2x-15)^5=(2x-15)^3

2) so saanh a)2^11 va 5^35

b)2^31 va 3^21

c)2^15 va 27^5.49^8 d)21^15 va 27^5.49^8

Answer 1

[Vũ Minh Tuấn]

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 \(x\in\left\{\frac{15}{2};8;7\right\}.\)

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