Trung bình cộng là :

( 43 + 42 + 38 ) : 3 = 41 ( học sinh )

Đáp số : 41 học sinh

a) Ta có:

\(0,\left(37\right)+0,\left(62\right)=0,\left(01\right)\cdot37+0,\left(01\right)\cdot62=\frac{1}{99}\cdot37+\frac{1}{99}\cdot62\)

\(=\frac{37}{99}+\frac{62}{99}=1\) (đpcm)

f) Ta có:

\(3^{n-1}+5\cdot3^{n-1}=162\)

\(\Rightarrow3^{n-1}\cdot\left(1+5\right)=162\)

\(\Rightarrow3^{n-1}\cdot6=27\cdot6\)

\(\Rightarrow3^{n-1}=27\)

\(\Rightarrow3^{n-1}=3^3\)

\(\Rightarrow n-1=3\)

\(\Rightarrow n=4\)

Vậy \(n=4\)

e) Ta có:

\(5^n+5^{n+2}=650\)

\(\Rightarrow5^n+5^n\cdot5^2=650\)

\(\Rightarrow5^n\cdot\left(1+5^2\right)=650\)

\(\Rightarrow5^n\cdot26=25\cdot26\)

\(\Rightarrow5^n=25\)

\(\Rightarrow5^n=5^2\)

\(\Rightarrow n=2\)

Vậy \(n=2\)

d) Ta có:

\(6^{3-n}=216\)

\(\Rightarrow6^{3-n}=6^3\)

\(\Rightarrow3-n=3\)

\(\Rightarrow n=3-3\)

\(\Rightarrow n=0\)

Vậy \(n=0\)\(\text{ }\)

c) Ta có:

\(\frac{64}{\left(-2\right)^n}=-32\)

\(\Rightarrow\frac{2^6}{\left(-2\right)^n}=\left(-2\right)^5\)

\(\Rightarrow2^n=2^6:\left(-2\right)^5\)

\(\Rightarrow2^n=2^6:2^5:\left(-1\right)\)

\(\Rightarrow2^n=2:\left(-1\right)\)

\(\Rightarrow2^n=-2\) (vô lí)

=> Không có giá trị n thỏa mãn.

Vậy không có giá trị n thỏa mãn.

\(\)

b) Ta có:

\(\frac{1}{2}\cdot2^n+4\cdot2^n=3^2\cdot2^5\)

\(\Rightarrow2^n\cdot\left(4+\frac{1}{2}\right)=9\cdot2^5\)

\(\Rightarrow2^n\cdot\frac{9}{2}=9\cdot2^5\)

\(\Rightarrow2^n=2^5\cdot9:\frac{9}{2}\)

\(\Rightarrow2^n=2^5\cdot2\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

Vậy \(n=6\)

a)

\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)

\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)

\(\Rightarrow9^n=3^n\)

\(\Rightarrow\left(3^2\right)^n=3^n\)

\(\Rightarrow3^{2n}=3^n\)

\(\Rightarrow2n=n\)

\(\Leftrightarrow n=0\)

Vậy \(n=0\)

Bài 1:

\(0,0\left(8\right)=\frac{1}{10}\cdot0,\left(8\right)=\frac{1}{10}\cdot0,\left(1\right)\cdot8=\frac{4}{5}\cdot\frac{1}{9}=\frac{4}{45}\)

\(0,1\left(2\right)=0,1+0,0\left(2\right)=\frac{1}{10}+\frac{1}{10}\cdot0,\left(2\right)=\frac{1}{10}+\frac{1}{10}\cdot0,\left(1\right)\cdot2=\frac{1}{10}+\frac{1}{5}\cdot\frac{1}{9}=\frac{1}{10}+\frac{1}{45}=\frac{11}{90}\)

\(0,1\left(23\right)=0,1+0,\left(23\right)=\frac{1}{10}+0,\left(01\right)\cdot23=\frac{1}{10}+\frac{1}{99}\cdot23=\frac{1}{10}+\frac{23}{99}=\frac{329}{990}\)

Bài 1:

\(5\cdot2\in Q\)

\(4,6351...\in I\)

\(-7,0903...\notin Q\)

\(1\cdot333\notin I\)

Bài 2:

+) Đúng

+) Đúng

+) Sai