a)  \(=\left(\frac{-1}{5}^3\right)^{100}va\left(\frac{-1}{3}^5\right)^{100}\)

\(=\left(\frac{-1}{125}\right)^{100}va\left(\frac{-1}{243}\right)^{100}\)

Mà \(\frac{-1}{125}>\frac{-1}{243}\)

\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b)\(2^{27}=8^9;3^{18}=9^9\)

Bài 1:

a: \(4\cdot64\cdot2^8\)

\(=2^2\cdot2^6\cdot2^8\)

\(=2^{2+6+8}=2^{16}\)

b: \(128\cdot27=2^7\cdot3^3\)

c: \(4\cdot27:\left(3^{11}\cdot\frac19\right)=4\cdot27:3^9=4\cdot\frac{3^3}{3^9}=\frac{4}{3^6}=\left(\frac{2}{3^3}\right)^2\)

Bài 2:

a: \(\left(\frac12\right)^3\cdot4+\frac34\)

\(=4\cdot\frac18+\frac34\)

\(=\frac48+\frac34=\frac48+\frac68=\frac{10}{8}=\frac54\)

b: Sửa đề: \(4^6\cdot\left(\frac12\right)^{12}\)

Ta có: \(4^6\cdot\left(\frac12\right)^{12}\)

\(=\left(2^2\right)^6\cdot\frac{1}{2^{12}}\)

\(=\frac{2^{12}}{2^{12}}=1\)

c: Ta có: \(\left(\frac12\right)^5-1,5^2\)

\(=\frac{1}{32}-\frac94\)

\(=\frac{1}{32}-\frac{72}{32}=\frac{-71}{32}\)

d: \(\frac{14^{16}\cdot35^7}{10^9\cdot7^{22}}\)

\(=\frac{7^{16}\cdot2^{16}\cdot7^7\cdot5^7}{2^9\cdot5^9\cdot7^{22}}\)

\(=\frac{7^{23}}{7^{22}}\cdot\frac{2^{16}}{2^9}\cdot\frac{5^7}{5^9}=7\cdot\frac{2^7}{5^2}=7\cdot\frac{128}{25}=\frac{896}{25}\)

1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)

=>3x-1/5=3/25 hoặc 3x-1/5=-3/25

=>3x=8/25 hoặc 3x=2/25

=>x=8/75 hoặc x=2/75

2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)

=>2x-1/3=2/9 hoặc 2x-1/3=-2/9

=>2x=5/9 hoặc 2x=1/9

=>x=5/18 hoặc x=1/18

Nhận xét:

Lũy thừa với số mũ chẵn của một số âm là một số dương

Lũy thừa với số mũ lẻ của một số âm là một số âm

neu so mu so lan chan thi la thua so duong

con neu so mu le lan thi thua so am

a: \(=2^2\cdot9\cdot\dfrac{1}{6\cdot9}\cdot\dfrac{4^2}{9^2}=\dfrac{2^2}{6}\cdot\dfrac{2^4}{3^4}=\dfrac{2^6}{2\cdot3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)

b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)

c: \(=\dfrac{\left(\dfrac{1}{2}\right)^3\cdot2^3\cdot\left(\dfrac{1}{2}\right)^2}{\left(-8\right)^2\cdot16}\cdot2^6=\dfrac{\dfrac{1}{2^2}}{64\cdot16}\cdot64=\dfrac{1}{4}:16=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\)

a: \(=2^2\cdot9\cdot\dfrac{1}{3^3\cdot2}\cdot\dfrac{2^4}{3^4}=\dfrac{2^4\cdot2^2}{2}\cdot\dfrac{9}{3^3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)

b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)

c: \(=\dfrac{\dfrac{1}{2^3}\cdot\dfrac{1}{2^2}\cdot8}{\left(-8\right)^2\cdot2^4}\cdot2^6=\dfrac{1}{2^2}\cdot2^6:2^{10}=\dfrac{2^4}{2^{10}}=\dfrac{1}{2^6}=\left(\dfrac{1}{8}\right)^2\)

D,       Vì 3^2=9 và -3^2=9 còn 5^2=25

D

`(1 1/4)^10 . (2/5)^20`

`=(5/4)^10 . (2/5)^20`

`=(5^10 .2^20)/(4^10 .5^20)`

`=(5^10 .4^10)/(4^10 .5^20)`

`=1/(5^10)`

`=(1/5)^10`

a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)

\(=-\dfrac{1}{10}\)

9<10

=>1/9>1/10

=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)

=>\(A>-\dfrac{1}{9}\)

b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

20<21

=>\(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)