3² > 3.2

nhớ cho tick!

3^2=9                                                                                                      3.2=6                                                                                                         3^2 > 3.2

3²=9;3.2=6

=>3²>3.2

Chắc mình phải lấy giấy vệ sinh thắt cổ tự tủ mất

Ta có: \({\left( {{{15}^3}} \right)^2}\) = 15³ . 15³ = 15³⁺³ = 15⁶

\({15^{3.2}}\) = 15⁶

Vậy \({\left( {{{15}^3}} \right)^2}\) = \({15^{3.2}}\)

\(\frac{4^2.5.11}{44.20}=\frac{2^4.5.11}{2^2.11.2^2.5}=\frac{2^4.5.11}{2^4.5.11}=1\)

\(\frac{13.15.6}{18.65.7}=\frac{13.3.5.2.3}{2.3^2.13.5.7}=\frac{3^2.2.5.13}{2.3^2.13.5.7}=\frac{1}{7}\)

\(\frac{4^2.5.11}{44.20};\frac{13.15.6}{18.65.7};\frac{7.2.8.5^2}{14.2.5}\)và\(\frac{2^3.3^2.5}{3.2^3.5^3}\)

Ta có : \(\frac{4^2.5.11}{44.20}=\frac{42.5.11}{4.11.4.5}=\frac{4^2.1.1}{4^2.1.1}=1\)

\(\frac{13.15.6}{18.65.7}=\frac{13.3.5.6}{3.6.15.5.7}=\frac{1.1.1.1}{1.1.1.1.7}=\frac{1}{7}\)

\(\frac{7.2.8.5}{14.2.5}=\frac{7.2.2^3.5^2}{2.7.2.5}=\frac{1.1.2^2.5}{1.1.1.1}=\frac{20}{1}\)

\(\frac{2^3.3^3.5}{3.2^3.5^3}=\frac{1.3^2.1}{1.1.5^2}=\frac{9}{25}\)

+) Ta có : \(7=7\)

\(25=5^2\)

\(MC=BCNN\left(7;25\right)=5^2.7=175\)

Khi đó : ................

p/s : kham khảo

`(4^2. 5.11)/(44.20)`

`=(4.11.4.5)/(4.11.4.5)`

`=1`

`(13.15.16)/(18.65.7)`

`=(13.15.16)/(2.3.3.13.5.7)`

`=8/21`

`(7.2.8.5^2)/(14.2.5)`

`=(14.2.4.5.5)/(14.2.5)`

`=4.5`

`=20`

`(2^3. 3^3. 5)/(3.2^3. 5^3)`

`=(2^3. 3.5.3^2)/(2^3. 3.5.5^2)`

`=(3^2)/(5^2)`

`=9/25`

**Quy đồng:

`(4^2. 5.11)/(44.20)=1=525/525`

`(13.15.16)/(18.65.7)=8/21=200/525`

`(7.2.8.5^2)/(14.2.5)=20=840/525`

`(2^3. 3^3. 5)/(3.2^3. 5^3)=9/25=189/525`

2³x5³=(2x5)³=10³

10⁴>10³ suy ra 10⁴>2³x5³

\(2^3\cdot5^3=\left(2\cdot5\right)^3=10^3\)   

Vì \(3< 4\Rightarrow10^3< 10^4\)   

Vậy \(2^3\cdot5^3< 10^4\)

(-2)^3.3^3.5^3.7.8/3.5^3.2^4.42

=(-2)^2.(-2).3^3.5^3.7.8/3.5^3.2^2.2^2.7.6

=4.(-2).3^3.5^3.7.8/3.5^3.4.4.7.6

=-1.3^2.8/6

=-72/6=-12

\(\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3.5^3.2^4.2.3.7}\)=\(\frac{\left(-2\right)^2.\left(-2\right).3^3.5^3.7.2^3}{3^2.5^3.2^5.7}\)=\(\frac{2^5.\left(-2\right).3}{2^5}\)=\(-6\)

Ta có:

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=\frac{-49}{50}\)