Question

so sánh các biểu thức sau:

a) A = \(\dfrac{13^{15}+1}{13^{16}+1}\) và B = \(\dfrac{13^{16}+1}{13^{17}+1}\)

b) A = \(\dfrac{1999^{1999}+1}{1999^{1998}+1}\) và B = \(\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

c) A = \(\dfrac{100^{100}+1}{100^{99}+1}\) và B = \(\dfrac{100^{69}+1}{100^{68}+1}\)

Answer 1

a) +) Có \(A=\frac{13^{15}+1}{13^{16}+1}\)=> 13A = \(\frac{13\left(13^{15}+1\right)}{13^{16}+1}\)

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

+) Có \(B=\frac{13^{16}+1}{13^{17}+1}\)=> 13B =\(\frac{13\left(13^{16}+1\right)}{13^{17}+1}\)

=\(\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)(2)

+) Từ (1) và (2) => \(1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

<=> 13A>13B <=> A> B

b) +) Có A=\(\frac{1999^{1999}+1}{1999^{1998}+1}\) => \(\frac{A}{1999}=\frac{1999^{1999}+1}{1999^{1999}+1999}=\frac{1999^{1999}+1999-1998}{1999^{1999}+1999}\)

=\(1-\frac{1998}{1999^{1999}+1999}\) (1)

+) Có B =\(\frac{1999^{2000}+1}{1999^{1999}+1}\)

=> \(\frac{B}{1999}=\frac{1999^{2000}+1}{1999^{2000}+1999}=1-\frac{1998}{1999^{2000}+1999}\)(2)

+) Từ (1) và (2) => \(1-\frac{1998}{1999^{1999}+1999}\)< \(1-\frac{1998}{1999^{2000}+1999}\)

<=> \(\frac{A}{1999}< \frac{B}{1999}\) <=> A< B

Answer 2

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

\(\dfrac{B}{10}=\dfrac{100^{69}+1}{100^{69}+10}=1-\dfrac{9}{100^{69}+10}\)

Ta có:  100^100+10>100^69+10

=>-9/(100^100+10)<-9/(100^69+10)

=>A/10<B/10

=>A<B