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\(\frac{3^{-m}}{81}=27\)
\(=>3^{-m}=27\cdot81\)
\(3^{-m}=2187\)
Vì nếu \(k^{-m}\) thì => k = \(\frac{1}{k^m}\)
mà 2187 \(\in N\)
=> Không tìm được m thỏa mãn yêu cầu đề bài.
\(\frac{3^{-m}}{81}=27\Rightarrow3^{-m}=27.81=2187=3^7\)
\(\Rightarrow-m=7\)
\(\Rightarrow m=-7\)
1. \(\frac{x^7}{81}=27\Leftrightarrow x^7=2187\)
\(\Leftrightarrow x^7=3^7\Leftrightarrow x=3\)
2. \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy,...
3.\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow x^8\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^8=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)
4. \(\left(3x-1\right)^3=\frac{-8}{27}\Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{9}\)
\(-1+\frac{1}{3}-\frac{1}{9}+\frac{1}{27}-\frac{1}{81}\)
\(-1+\frac{2}{9}+\frac{2}{81}\)
\(-1+\frac{20}{81}\)
Tự tính nhé
Chúc bạn học tốt
3^12.(3^4)^11/(3^3)^10.(3^2)^15=3^12.3^44/3^30.3^30=3^56/3^30
=3^17.9^22/3^30.9^15=9^7/3^13=3^14/3^13=3^1=3
giùm nhé bạn
Bài 1
A= \(\frac{81^{10}.3^{17}}{27^{10}.9^{13}}\)
= \(\frac{\left(3^4\right)^{10}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{13}}\)
= \(\frac{3^{40}.3^{17}}{3^{30}.3^{26}}\)
= \(\frac{3^{57}}{3^{56}}\)= 3
\(\frac{x}{7}\)= \(\frac{-12}{49}\)
=> 49x = (-12) x 7
=> 49x = -84
=> x= \(\frac{-12}{7}\)
\(\frac{2^{15}.9^4}{6^3.8^3}\)=\(\frac{2^{15}.\left(3^2\right)^3}{\left(2.3\right)^3.\left(2^3\right)^3}\)=\(\frac{2^{15}.3^6}{2^3.3^3.2^9}\)=\(\frac{2^{15}.3^6}{2^{12}.3^3}\)=\(2^3.3^3\)=8.27=216
\(\frac{3^{-m}}{81}=27\)
\(\frac{3^{-m}}{3^4}=3^3\)
\(3^{-m}=3^3\times3^4\)
\(3^{-m}=3^7\)
\(-m=7\)
\(m=7\)
m = -7 nhể?