Question

[Bùi Vân Giang]

So sánh các phân số sau:

a)\(\frac{9^{10}+1}{9^{11}+1};\frac{9^{11}+1}{9^{12}+1}\)                                                                       b)\(\frac{11^{221}+1}{11^{222}+1};\frac{11^{222+1}}{11^{223}+1}\)

c)\(\frac{2000^{2002}+1}{2000^{2001}+1};\frac{2000^{2003}+1}{2002^{2002}+1}\)                                         d)\(\frac{10^{20}+9}{10^{20}-6};\frac{10^{21}+5}{10^{21}+5}\)

Giúp mk với , thanks nhiều

Answer 1

[Trí Tiên亗]

Mình làm mẫu câu a) các câu sau tương tự nhé :

Đăt \(A=\frac{9^{10}+1}{9^{11}+1}\Rightarrow9A=\frac{9^{11}+9}{9^{11}+1}=1+\frac{8}{9^{11}+1}\)

\(B=\frac{9^{11}+1}{9^{12}+1}\Rightarrow9B=\frac{9^{12}+9}{9^{12}+1}=1+\frac{8}{9^{12}+1}\)

Ta có : \(9^{11}+1< 9^{12}+1\)

\(\Rightarrow\frac{8}{9^{11}+1}>\frac{8}{9^{12}+1}\)

\(\Rightarrow1+\frac{8}{9^{11}+1}>1+\frac{8}{9^{12}+1}\)

\(\Rightarrow9A>9B\)

hay : \(A>B\)

Vậy : \(\frac{9^{10}+1}{9^{11}+1}>\frac{9^{11}+1}{9^{12}+1}\)

Answer 2

[•๖ۣۜHυүềη ²ƙ⁷ ( ๖ۣۜTεαм ๖ۣۜBạ¢ɦ ๖ۣۜ...]

Đặt \(\frac{9^{10}+1}{9^{11}+1}\)là A

\(\frac{9^{11}+1}{9^{12}+1}\) là B

\(\Rightarrow9A=\frac{9^{11}+9}{9^{11}+1}=1+\frac{8}{9^{11}+1}\)

\(\text{​​}\Rightarrow9B=\frac{9^{12}+9}{9^{12}+1}=1+\frac{8}{9^{12}+1}\)

\(\text{Vì }\frac{8}{9^{11}+1}>\frac{8}{9^{12}+1}\)

\(\Rightarrow9A>9B\Rightarrow A>B\)

các bài khác cũng tương tự nhé nhé