3¹¹<5⁸

Ai k mk mk k lai

3¹¹<5⁸

Đúng đó

a: \(5^{300}=25^{150}\)

\(3^{450}=27^{150}\)

mà 25<27

nên \(5^{300}< 3^{450}\)

a: 5300=251505300=25150

3450=271503450=27150

mà 25<27

nên 5300<3450

(x-3)^11=(x-3)^7

(x-3)^11-(x-3)^7=0

(x-3)^7[(x-3)^4-1)]=0

\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^7=0\\\left(x-3\right)^4-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^4=1\end{cases}}\)\(\Rightarrow\)x=3; x=2; x=4

Vậy x=3 hoặc x=2 hoặc x=4

Ta có (x-3)^11 = (x-3)^7

  <=>  \(\hept{\begin{cases}x-3=0\\x-3=1\\x-3=-1\end{cases}}\)

   <=> \(\hept{\begin{cases}x=3\\x=4\\x=2\end{cases}}\)

Mình ra x=2;3;4 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)

mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)

\(\frac{11}{15}=\frac{15-4}{15}=1-\frac{4}{15}\)

\(\frac{145}{149}=\frac{149-4}{149}=1-\frac{4}{149}\)

Có \(\frac{4}{15}>\frac{4}{149}\)suy ra \(\frac{11}{15}< \frac{145}{149}\).

a: 4/5=1-1/5

5/6=1-1/6

mà 1/5>1/6

nên 4/5<5/6

b: 3/7=9/21>5/21

c: 13/12<14/12=7/6