Ta có:

\(M=\dfrac{100^{100}+1}{100^{99}+1}\)

\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}\)

\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}\)

\(\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}\) 

\(N=\dfrac{100^{101}+1}{100^{100}+1}\)

\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}\)

\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}\)

\(\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}\)

Mà: \(100^{101}>100^{100}\)

\(\Rightarrow100^{101}+100>100^{100}+100\)

\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)

\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)

\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)

\(\Rightarrow N< M\)

vì A và B đều có 1 nên ta bỏ 1 đi

Ta có : 100^100-100^99=9000......00000( tổng cộng có 198 số 0)

\(\frac{1}{100^{98}}=\frac{100}{100^{99}}\)nên \(\frac{1}{100^{99}}-\frac{1}{100^{98}}=\frac{-99}{100^{99}}\)

nhưng 900....000( 198 số 0) lớn hơn \(\frac{-99}{100^{99}}\)

=>A>B

A=100^101+1/100^100+1

B=100^100+1/100^99+1

A<100^101+1+99/100^100+1+99

A<100^101+100/100^100+100

A<100.(100^100+1)/100.(100^99+1)

A<100^100+1/100^99+1=B

=> A<B

Vậy A<B

C>D, chắc chắn đó

Ta có\(10A=\frac{100^{100}+1}{100^{100}+10}=\frac{100^{100}+1}{100^{100}+1+9}=\frac{100^{100}+1}{1+9}\)

\(10B=\frac{100^{98}+1}{100^{98}+10}=\frac{100^{98}+1}{100^{98}+1+9}=\frac{100^{98}+1}{1+9}\)

Vì\(\frac{100^{100}+1}{1+9}>\frac{100^9+1}{1+9}\)

=>10A>10B

=>A>B