Question

So sánh

A=\(\frac{100^{100}+1}{100^{90}+1}\)và B=\(\frac{100^{99}+1}{100^{89}+1}\)

Answer 1

A = \(\frac{100^{100}+1}{100^{90}+1}\)

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

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

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

B = \(\frac{100^{99}+1}{100^{89}+1}\)

\(\frac{1}{100^{10}}B=\frac{100^{99}+1}{100^{99}+100^{10}}\)

\(\frac{1}{100^{10}}B=\frac{100^{99}+100^{10}-100^{10}+1}{100^{99}+100^{10}}\)

\(\frac{1}{100^{10}}B=1+\frac{-100^{10}+1}{100^{99}+100^{10}}\)

Vì \(\frac{-100^{10}+1}{100^{100}+100^{10}}< \frac{-100^{10}+1}{100^{99}+10^{10}}\)nên A < B