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a) 32 . 3n = 35
=> 3n = 35 : 32
=> 3n = 33
=> n = 3
b) (22 : 4) . 2n = 4
=> (4 : 4) . 2n = 4
=> 2n = 4
=> 2n = 22
=> n = 2
c) \(\frac{1}{9}.3^4.3^n=3^7\)
\(\Rightarrow3^{-2}.3^4.3^n=3^7\)
\(\Rightarrow3^{-2+4+n}=3^7\)
\(\Rightarrow3^{2+n}=3^7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=5\)
d) \(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=n\)
\(\Rightarrow3^{-2+3n}=n\)
\(\Rightarrow-2+3n=n\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=1\)
1)\(\frac{11\cdot3^{29}-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
2)\(|2x-3|+2^3\cdot3=25\Rightarrow x=1;2\)
3) \(x183y=61831\Rightarrow x=6;y=1\)
4)\(B=\frac{n-1}{n-4}\Rightarrow n=1;3;5;7\)
5)\(\left(2x+1\right)\cdot\left(y^2-5\right)=12\Rightarrow x=1;y=3\)
mình là người đúng nhất ở bài 3 vì 61831 mới chia 2,5,9 dư 1
k cho mình nhé
a ) \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}\)
\(=\frac{3^{28}\left(11\cdot3-3^2\right)}{2^2\cdot3^{28}}=\frac{33-9}{4}=\frac{24}{4}=6\)
b ) \(27^{16}\div9^{10}\)
\(=\left(3^3\right)^{16}\div\left(3^2\right)^{10}\)
\(=3^{48}\div3^{20}\)
\(=3^{48-20}\)
\(=3^{28}\)
Bg
a) \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
= \(\frac{11.3^{22}.3^7-3^{30}}{2^2.3^{28}}\)
= \(\frac{11.3^{29}-3.3^{29}}{4.3^{28}}\)
= \(\frac{8.3^{29}}{4.3^{28}}\)
= \(\frac{2.3}{1.1}\)
= 6
b) 2716 ÷ 910
= 33.16 ÷ 32.10
= 348 ÷ 320
= 348 - 20
= 328
Đề sai thì phải ! Học Lớp 7 mới giải xong bài này !
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{9}\cdot\left(3^3\right)^n=3^n\)
\(\frac{1}{9}\cdot3^{3n}=3^n\)
\(\frac{1}{9}=3^n\text{ : }3^{3n}\)
\(\frac{1}{9}=3^{-2n}\)
\(\frac{1}{3^2}=\frac{1}{3^{2n}}\)
\(\Rightarrow\text{ }3^{2n}=3^2\)
\(3^{2n}-3^2=0\)
\(3\left(3^{2n-1}-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3=0\text{ ( Vô lí ) }\\3^{2n-1}-3=0\end{cases}}\) \(\Rightarrow\text{ }3^{2n-1}=3\) \(\Rightarrow\text{ }2n-1=1\) \(\Rightarrow\text{ }2n=2\) \(\Rightarrow\text{ }n=1\)
Vậy \(n=1\)
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\frac{3^{3n}}{3^2}=3^n\)
\(3^{3n}=3^2\cdot3^n\)
\(3^{3n}=3^{n+2}\)
\(\Rightarrow\text{ }3n=n+2\)
\(3n-n=2\)
\(2n=2\)
\(n=2\text{ : }2\)
\(n=1\)