\(M=\dfrac{10^{20}+1}{10^{19}+1}\)

\(N=\dfrac{10^{21}+1}{10^{20}+1}< \dfrac{10^{21}+1+9}{10^{20}+1+9}=\dfrac{10^{21}+10}{10^{20}+10}=\dfrac{10\left(10^{20}+1\right)}{10\left(10^{19}+1\right)}=\dfrac{10^{20}+1}{10^{19}+1}=M\)

\(\Rightarrow N< M\)

Sai rồi

Giúp với

Chứng minh nếu a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)

Do a/b < 1 => a < b

=> am < bm

=> am + ab < bm + ab

=> a.(b+m) < b.(a+m)

=> a/b < a+m/b+m

Áp dụng điều trên ta có: B = 10²⁰ + 1/ 10²¹ + 1 < 1

=> B < 10²⁰ + 1 + 9/10²¹ + 1 + 9

=> B < 10²⁰ + 10/10²¹ + 10

=> B < 10.(10¹⁹ + 1)/10.(10²⁰ + 1)

=> B < 10¹⁹+1/10²⁰+1 = A

=> B < A

b) n + 1 chia hết cho n - 2

=> n - 2 + 3 chia hết cho n - 2

Do n - 2 chia hết cho n - 2

=> 3 chia hết cho n - 2

=> n - 2 thuộc { 1 ; -1 ; 3 ; -3}

=> n thuộc { 3 ; 1 ; 5 ; -1}

Vậy n thuộc { 3 ; 1 ; 5 ; -1}

ehfhthjjm458855 ngu

Ta có : 

\(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}\)

Vậy \(\frac{10^{19}+1}{10^{20}+1}>\frac{10^{20}+1}{10^{21}+1}\)

Áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\) (a;b;c \(\in\) N*)

Ta có:

\(B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}\)

\(B< \frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

=> A > B

Ta dùng bất đẳng thức\(\frac{a}{b}<\frac{a+n}{b+n}\left(n\ne0\right)\)

Ta có \(B=\frac{10^{20}+1}{10^{21}+1}<\frac{10^{20}+1+9}{10^{21}+1+9}<\frac{10^{20}+10}{10^{21}+10}<\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\) 

               \(<\frac{10^{19}+1}{10^{20}+1}\)

Vậy \(A>B\)

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a=b moi dung chu

Ta có:\(B=\frac{10^{20}+1}{10^{21}+1}< 1\Rightarrow B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

=> A > B

đúng rồi mk cũng làm như vậy

\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)

NÊN \(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

VẬY B<A