1/ \(\left(\frac{3}{7}\right)^n=\frac{81}{2401}\)
\(\Rightarrow\left(\frac{3}{7}\right)^n=\left(\frac{3}{7}\right)^4\)
\(\Rightarrow n=4\)
Bài 1:
1. \(\left(\frac{3}{7}\right)^n=\frac{81}{2401}\)
⇒ \(\left(\frac{3}{7}\right)^n=\left(\frac{3}{7}\right)^4\)
⇒ \(n=4\)
Vậy \(n=4.\)
2. \(x^5=x^3\)
⇒ \(x^5-x^3=0\)
⇒ \(x^3.\left(x^2-1\right)=0\)
⇒ \(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=0\\x^2=0+1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;-1\right\}.\)
3. \(\left(x-\frac{4}{11}\right)^3=343\)
⇒ \(\left(x-\frac{4}{11}\right)^3=7^3\)
⇒ \(x-\frac{4}{11}=7\)
⇒ \(x=7+\frac{4}{11}\)
⇒ \(x=\frac{81}{11}\)
Vậy \(x=\frac{81}{11}.\)
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2/ \(x^5=x^3\)
\(\Rightarrow x=0\) hoặc \(x=1\)
Vì \(0^5=0^3\)
\(1^5=1^3\)
3/ \(\left(x-\frac{4}{11}\right)^3=343\)
\(\left(x-\frac{4}{11}\right)^3=7^3\)
=> \(x-\frac{4}{11}=7\)
\(x=7+\frac{4}{11}\)
\(x=\frac{77}{11}+\frac{4}{11}=\frac{81}{11}\)