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18 tháng 5 2021

\(a.\)

\(A=\)\(\frac{10^{15}+1}{10^{16}+1}\)

\(10A=\) \(\frac{10\left(10^{15}+1\right)}{10^{16}+1}\)

\(10A=\) \(\frac{10^{16}+10}{10^{16}+1}\)

\(10A=\)\(\frac{10^{16}+1+9}{10^{16}+1}\)

\(10A=\frac{10^{16}+1}{10^{16}+1}+\frac{9}{10^{16}+1}\)

\(10A=1+\frac{9}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

\(10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}\)

\(10B=\frac{10^{17}+1+9}{10^{17}+1}\)

\(10B=\frac{10^{17}+1}{10^{17}+1}+\frac{9}{10^{17}+1}\)

\(10B=1+\frac{9}{10^{17}+1}\)

\(\Rightarrow10B< 10A\Rightarrow B< A\)\(\text{( vì tự làm ) }\)

19 tháng 5 2021

xin lỗi hôm qua mk đang làm thì phải đy học zoom học xong quên h mới nhơ ra làm típ :)

\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(B=\frac{3}{8^4}+\frac{7}{8^3}=\frac{3}{8^4}+\frac{3}{8^3}+\frac{4}{8^3}\)

Vì \(\frac{4}{8^4}< \frac{4}{8^3}\)=.> A < B

13 tháng 4 2018

\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

Nhận thấy: \(\frac{9}{10^{17}+1}< \frac{9}{10^{16}+1}\)=> 10B < 10A

=> A > B

25 tháng 3 2019

A = ( 10^15+1 ) / ( 10^16+1 ) => 10A = ( 10^16+10 ) / ( 10^16+1 ) = 1 + ( 9/10^15+1 )

B = ( 10^16+1 ) / ( 10^17+1 ) => 10B = ( 10^17+10 ) / ( 10^17+1 ) = 1 + ( 9/10^16+1 )

Vì 10^15+1 < 10^16+1 nên 9/10^15+1 > 9/10^16+1 => 1 + ( 9/10^15+1 ) > 1 + ( 9/10^16+1 )

Vậy A > B

7 tháng 3 2017

Ta có :

\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

Vì \(10^{16}+1< 10^{17}+1\) nên \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) \(\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)

=> 10A > 10B Do đó A > B

Vậy A > B

7 tháng 3 2017

\(A=\frac{10^{15}+1}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

Ta có:

\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{\left(10^{15}+1\right).10}{\left(10^{16}+1\right).10}=\frac{10^{16}+10}{10^{17}+10}=\frac{10^{16}+1+9}{10^{17}+1+9}\)

\(B=\frac{10^{16}+1}{10^{17}+1}< 1\)

\(\Rightarrow B=\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}=A\)

Vậy B < A

Bài 1:

Ta có:

\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)

\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)

Lại có:

\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)

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

Bài 2:

Ta có:

\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

\(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)

\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\Rightarrow A>B\)

8 tháng 2 2017

\(A=\frac{10^{15}+1}{10^{16}+1}\)

\(\Rightarrow10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)

\(A=\frac{10^{16}+1}{10^{17}+1}\)

\(\Rightarrow10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

Vì \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\left(Do10^{16}+1< 10^{17}+1\right)\)

\(\Rightarrow10A>10B\)

\(\Rightarrow A>B\)

17 tháng 3 2015

tính 10A và  10B rồi so sánh

 

25 tháng 3 2019

sai đề rồi bạn.\(\frac{a}{b}>\frac{a+c}{b+c}\) với \(a>b\) mới đúng nha.

Ta có:\(A=\frac{10^{17}+1}{10^{16}+1}>\frac{10^{17}+1+9}{10^{16}+1+9}=\frac{10^{17}+10}{10^{16}+10}=\frac{10\left(10^{16}+1\right)}{10\left(10^{15}+1\right)}=\frac{10^{16}+1}{10^{15}+1}\)

\(\Rightarrow A>B\)

25 tháng 3 2019

:DDDDDD

17 tháng 8 2017

Ta có:

\(A=\frac{10^{15}+1}{10^{16}+1}\)

\(10A=\frac{10^{16}+10}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}\)

Ta so sánh \(10A\) và \(10B\)

Có: 

\(10A:\) Mẫu - tử = 9

\(10B:\) Mẫu - tử = 9

Lại có:

 \(\frac{10^{16}+10}{10^{16}+1}\) \(-1\)\(=\frac{9}{10^{16}+1}\)

\(\frac{10^{17}+10}{10^{17}+1}-1=\frac{9}{10^{17}+1}\)

Vì \(\frac{9}{10^{16}+1}\)\(>\frac{9}{10^{17}+1}\)nên \(10A>10B\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

17 tháng 8 2017

Theo bải ra ta có:

A=\(\frac{10^{15}+1}{10^{16}+1}\)=> 10A =.\(\frac{10.\left(10^{15}+1\right)}{10^{16}+1}\)\(\frac{10.10^{15}+1.10}{10^{16}+1}\)

                                      = \(\frac{10.10^{15}+10}{10^{16}+1}\)=\(\frac{10^{16}+1+9}{10^{16}+1}\)\(1+\frac{9}{10^{16}+1}\)

B= \(\frac{10^{16}+1}{10^{17}+1}\)=> 10B = \(\frac{10.\left(10^{16}+1\right)}{10^{17}+1}\)=\(\frac{10.10^{16}+1.10}{10^{17}+1}\)

                                       = \(\frac{10.10^{16}+10}{10^{17}+1}\)\(\frac{10^{17}+1+9}{10^{17}+1}\)\(1+\frac{9}{10^{17}+1}\)

Vì 1=1 mà \(\frac{9}{10^{16}+1}\)>   \(\frac{9}{10^{17}+1}\)nên => 10A > 10B => A>B

Vậy A>B.