`A=(20^10+1)/(20^11+1)`

`=>20A=(20^11+20)/(20^11+1)=1+19/(20^11+1)`

Hoàn toàn tương tự: `20B=1+19/(20^12+1)`

Vì `19/(20^12+1)<19/(20^11+1)`

`=>20B<20A`

`=>B<A`

Tối nhé, mình bận,

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Em cố gắng gõ latex nha

a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{2}{3}\)

b: \(=\dfrac{11^9\cdot3^{18}}{3^{18}\cdot11^{11}}=\dfrac{1}{121}\)

a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{20}{30}=\dfrac{2}{3}\)

b: \(=\dfrac{11^9\cdot3^{18}}{3^{10}\cdot11^{11}\cdot3^8}=\dfrac{1}{121}\)

Ta có : 

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-..-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{52.55}\right)=\frac{3}{11}\)

Đặt \(A=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)

\(A=10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(A=10.\frac{4}{55}=\frac{40}{55}=\frac{8}{11}\)

=> \(x-\frac{8}{11}=\frac{3}{11}\)

\(x=\frac{3}{11}+\frac{8}{11}=\frac{11}{11}=1\)

Ủng hộ mk nha !!! ^_^

 I don't no