Question

So sánh:A=\(\dfrac{10^{100}+1}{10^{99} +1}\) và B=\(\dfrac{10^{101}+1}{10^{100}+1}\)

Answer 1

ta có:

1/10.A=10¹⁰⁰+1/10(10⁹⁹+1)

1/10.A=10¹⁰⁰+1/10¹⁰⁰+10

1/10.A=1-(9/10¹⁰⁰+10)

 

1/10.B=10¹⁰¹+1/10(10¹⁰⁰+1)

1/10.B=10¹⁰¹+1/10¹⁰¹+10

1/10.B=1-(9/10¹⁰¹+10)

vì(10¹⁰¹+10)>(10¹⁰⁰+1)=>  9/10¹⁰¹+10 < 9/10¹⁰⁰+10 => 1-(9/10¹⁰¹+10) > 1-(9/10¹⁰⁰+10)

hay 1/10.A>1/10.B

=>A>B

Answer 2

ta có:

1/10.A=10100+1/10(1099+1)

1/10.A=10100+1/10100+10

1/10.A=1-(9/10100+10)

1/10.B=10101+1/10(10100+1)

1/10.B=10101+1/10101+10

1/10.B=1-(9/10101+10)

vì(10101+10)>(10100+1)=>  9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)

hay 1/10.A<1/10.B

=>A<B

Answer 3

Đáp án dưới mới đúng nhé 

vừa mình làm nhầm

Answer 4

\(A=\dfrac{10^{100}+1}{10^{99}+1}\)

\(\dfrac{1}{10}A=\dfrac{10^{100}+1}{10^{100}+10}=\dfrac{10^{100}+10-9}{10^{100}+10}\)

\(\dfrac{1}{10}A=\dfrac{10^{100}+10}{10^{100}+10}-\dfrac{9}{10^{100}+10}\)

\(\dfrac{1}{10}A=1-\dfrac{9}{10^{100}+10}\)

\(B=\dfrac{10^{101}+1}{10^{100}+1}\)

\(\dfrac{1}{10}B=\dfrac{10^{101}+1}{10^{101}+10}=\dfrac{10^{101}+10-9}{10^{101}+10}\)

\(\dfrac{1}{10}B=\dfrac{10^{101}+10}{10^{101}+10}-\dfrac{9}{10^{101}+10}\)

\(\dfrac{1}{10}B=1-\dfrac{9}{10^{101}+10}\)

Vì \(10^{100}+10< 10^{101}+10\) nên \(\dfrac{9}{10^{100}+10}>\dfrac{9}{10^{101}+10}\)

Vậy \(1-\dfrac{9}{10^{100}+10}< 1-\dfrac{9}{10^{101}+100}\) hay \(A< B\)

Answer 5

\(\frac{1}{10}\) \(A\) \(=\) \(\) \(\frac{10^{100}+1}{10^{100}+10}\) \(=\) \(\frac{10^{100}+10-9}{10^{100}+10}=1-\frac{9}{10^{100}+10}\)

\(\frac{1}{10}B\) \(=\) \(\frac{10^{101}+1}{10^{101}+10}=\frac{10^{101}+10-9}{10^{101}+10}=1-\frac{9}{101^{101}+10}\)

Vì \(1-\frac{9}{10^{101}+10}>1-\frac{9}{10^{100}+10}\) nên \(\frac{1}{10}A<\frac{1}{10}B\)

Suy ra \(A

Vậy \(A

Answer 6

ta có công thức\(\frac{a^2+m}{b^1+m}>\frac{c^3+m}{d^2+m}\) ;\(\frac{a^2}{b^1}>\frac{c^3}{d^2}\)

dựa vào công thức thứ 2 ta có:A>B