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

\(\Rightarrow10A>10B\)

\(\Rightarrow A>B\)

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

18 tháng 12 2017

giúp mình với mai phải nộp rồi

2 tháng 2 2015

Ta có:

10A=1016+10/1016+1=1+​​(9/1016+1)

10B=1017+10/1017+1=1+(9/1017+1)

Vì 9/1016+1 > 9/1017+1 nên 10A>10B,do đó A>B

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.

8 tháng 5 2016

\(\frac{10^{15}+1}{10^{16}+1}=\frac{10^{16}+10}{10^{17}+10}\)

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

Vậy B<A

8 tháng 5 2016

Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\) ; \(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

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

8 tháng 3 2015

c1: Ta có \(\frac{10^{15}+1+9}{10^{16}+1+9}\)=\(\frac{10^{15}+10}{10^{16}+10}\)=\(\frac{10\left(10^{14}+1\right)}{10\left(10^{15}+1\right)}\)=\(\frac{10^{14}+1}{10^{15}+1}\)

Vì \(\frac{10^{15}+1}{10^{16}+1}\)<1 nên \(\frac{10^{15}+1+9}{10^{16}+1+9}\)>\(\frac{10^{15}+1}{10^{16}+1}\)Vậy A<B

1 tháng 4 2019

c2 Ta có 10A=\(\frac{10^{16+10}}{10^{16}+1}\)\(\frac{10^{16}+1+9}{10^{16+1}}\)=\(\frac{10^{16}+1}{10^{16}+1}\)+  \(\frac{9}{10^{16}+1}\)=1+\(\frac{9}{10^{16}+1}\)

Tương tự câu B cung nhân vs 10 đc 1 +\(\frac{9}{10^{17}+1}\) vì 1+\(\frac{9}{10^{17}+1}\)< 1 + \(\frac{9}{10^{16}+1}\)\(\Rightarrow\)10B<10A\(\Rightarrow\)B<A

Vậy A>B 

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