Question

\(C=\frac{100^{16}+1}{100^{17}+1}\)và \(D=\frac{100^{15}+1}{100^{16}+1}\)

Answer 1

Ta có : 

\(100C=\frac{100^{17}+100}{100^{17}+1}=\frac{100^{17}+1+99}{100^{17}+1}=\frac{100^{17}+1}{100^{17}+1}+\frac{99}{100^{17}+1}=1+\frac{99}{100^{17}+1}\)

\(100D=\frac{100^{16}+100}{100^{16}+1}=\frac{100^{16}+1+99}{100^{16}+1}=\frac{100^{16}+1}{100^{16}+1}+\frac{99}{100^{16}+1}=1+\frac{99}{100^{16}+1}\)

Vì \(\frac{99}{100^{17}+1}< \frac{99}{100^{16}+1}\) nên \(1+\frac{99}{100^{17}+1}< 1+\frac{99}{100^{16}+1}\) hay \(100A< 100B\)

\(\Rightarrow\)\(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Answer 2

Ta có : \(100C=\frac{100^{17}+100}{100^{17}+1}=1+\frac{100}{100^{17}+1}\)

         \(100D=\frac{100^{16}+100}{100^{16}+1}=1+\frac{100}{100^{16}+1}\)

Mà \(\frac{100}{100^{17}+1}< \frac{100}{100^{16}+1}\)

\(\Rightarrow10C< 10D\Rightarrow C< D\)