\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5=\left(\dfrac{10}{10000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)

 \(\dfrac{10}{10000}< \dfrac{81}{10000}\)

\(\Rightarrow\left(\dfrac{10}{10000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)

\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)

Ta có:

\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)

Ta thấy: \(\dfrac{1}{1000}< \dfrac{81}{10000}\)

\(\Rightarrow\left(\dfrac{1}{1000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)

\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)

 \(\left(\dfrac{3}{10}\right)^{20}=3^{20}.\left(\dfrac{1}{10}\right)^{20}\)

\(=3^{20}.\left(\dfrac{1}{10}\right)^5.\left(\dfrac{1}{10}\right)^{15}\)

\(=\dfrac{3^{20}}{10^5}.\left(\dfrac{1}{10}\right)^{15}\)

\(=\left(\dfrac{81}{5}\right)^5.\left(\dfrac{1}{10}\right)^{15}\)

Mà \(\left(\dfrac{1}{10}\right)^{15}=\left(\dfrac{1}{10}\right)^{15}\)

\(\Rightarrow\left(\dfrac{3}{10}\right)^{20}>\left(\dfrac{1}{10}\right)^{15}\)

Ta có:

\(\left(\dfrac{1}{10}\right)^{15}=\left(\left(\dfrac{1}{10}\right)^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left(\left(\dfrac{3}{10}\right)^4\right)^5=\left(\dfrac{81}{10000}\right)^5\)

Ta có: \(\left(\dfrac{1}{10}\right)^{15}=\left(\dfrac{1}{10}^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left(\dfrac{3}{10}^4\right)^5=\left(\dfrac{3}{10000}\right)^5\)

Vì \(\dfrac{1}{1000}>\dfrac{3}{10000}\) nên \(\left(\dfrac{1}{10}\right)^{15}>\left(\dfrac{3}{10}\right)^{20}\)

>

Trl:

Đây ko phải là bài lp 5 bn nhé.

Hok tốt!

trời đất toán lớp 5 khó bằng toán 6 lun á

đây ko phải lop5

Giải:

Ta có:

A=20¹⁰+1/20¹⁰-1

A=20¹⁰-1+2/20¹⁰-1

A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-1<2/20¹⁰-3 nên A<B

Chúc bạn học tốt!

Lời giải:

$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}$

$B=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}$

Vì $20^{10}-1> 20^{10}-3$

$\Rightarrow \frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}$

$\Rightarrow 1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}$

$\Rightarrow A< B$

\(A=\frac{2010+1}{2010-1}\)

\(A=1+\frac{2}{2010-1}>1\)

\(B=\frac{2010-1}{2010-3}\)

\(B=1-\frac{2}{2010-3}<1\)

Từ đó A > B

Ta thấy : A =\(\frac{20^{10}+1}{20^{10}-1}>1\) 
Ta có : A=\(\frac{20^{10}+1}{20^{10}-1}>\frac{20^{10}+1-2}{20^{10}-1-2}=\frac{20^{10}-1}{20^{10}-3}=B\)
Vậy A > B

cám ơn bạn

Theo đề ta có:

A= 20^10+1/20^10-1 (1)

Từ (1) ta có: 20^10+1/20^10-1>20^10+1-2/20^10-1-2=20^10-1/20^10-3=B.

Vậy A>B

Ta thấy:\(A=\frac{20^{10}+1}{20^{10}-1}>1\)

Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}>\frac{20^{10}+1-2}{20^{10}-1-2}=\frac{20^{10}-1}{20^{10}-3}=B\)

Vậy \(A>B\)

Ta có:

\(A=\frac{20^{10}+1}{20^{10}-1}\)

\(=\frac{20^{10}-1+2}{20^{10}-1}\)

\(=1+\frac{2}{20^{10}-1}\)

\(B=\frac{20^{10}-1}{20^{10}-3}\)

\(=\frac{20^{10}-3+2}{20^{10}-3}\)

\(=1+\frac{2}{20^{10}-3}\)

Ta lại có:

\(20^{10}-1>20^{10}-3\)

\(\Rightarrow\)\(\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}\)

\(\Rightarrow\)\(1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}\)

Vậy ta kết luận A < B