Question

[Long123]

So sánh :

A = 1001¹⁰⁰¹/1002¹⁰⁰²     và

B = 1001¹⁰⁰¹+ 101101/1002¹⁰⁰²+101202

Answer 1

[Nguyễn Linh Chi]

\(A=\frac{1001^{1001}}{1002^{1002}}=\frac{1001^{1000}.1001}{1002^{1001}.1002}\)

\(B=\frac{1001^{1001}+101101}{1002^{1002}+101202}=\frac{1001.1001^{1000}+1001.101}{1002.1002^{1001}+1002.101}\)

\(=\frac{1001\left(1001^{1000}+101\right)}{1002\left(1002^{1001}+101\right)}\)

Xét \(\frac{1001^{1000}+101}{1002^{1001}+101}\)\(-\frac{1001^{1000}}{1002^{1001}}\)

\(=\frac{1002^{1001}\left(1001^{1000}+101\right)-1001^{1000}\left(1002^{1001}+101\right)}{\left(1002^{1001}+101\right).1002^{1001}}\)

\(=\frac{1002^{1001}.1001^{1000}+1002^{1001}.101-1001^{1000}.1002^{1001}-1001^{1000}.101}{\left(1002^{1001}+101\right).1002^{1001}}\)

\(=\frac{101\left(1002^{1001}-1001^{1000}\right)}{\left(1002^{1001}+101\right).1002^{1001}}>0\)

=> \(\frac{1001^{1000}+101}{1002^{1001}+101}\)\(>\frac{1001^{1000}}{1002^{1001}}\)

=> \(\frac{1001\left(1001^{1000}+101\right)}{1002\left(1002^{1001}+101\right)}>\frac{1001^{1000}.1001}{1002^{1001}.1002}\)

=> \(B>A\)

Answer 2

[Long123]

Mình cảm ơn ạ! Hi vọng sau này ban sẽ giúp mình nữa nha ^^