Toán CMR A= 1/11+1/12+1/13+1/14+…+1/40.1 < A < 2 11/09/2021 By Parker CMR A= 1/11+1/12+1/13+1/14+…+1/40.1 < A < 2
Giải thích các bước giải: Ta có: $A=\dfrac{1}{11}+\dfrac1{12}+\dfrac1{13}+…+\dfrac1{40}$ $\to A=(\dfrac1{11}+\dfrac1{12}+…+\dfrac1{20})+(\dfrac1{21}+\dfrac1{22}+…+\dfrac1{40})$ $\to A>(\dfrac1{20}+\dfrac1{20}+…+\dfrac1{20})+(\dfrac1{40}+\dfrac1{40}+…+\dfrac1{40})$ $\to A>\dfrac12+\dfrac12$ $\to A>1$ Ta có: $A=\dfrac{1}{11}+\dfrac1{12}+\dfrac1{13}+…+\dfrac1{40}$ $\to A=(\dfrac1{11}+\dfrac1{12}+…+\dfrac1{20})+(\dfrac1{21}+\dfrac1{22}+…+\dfrac1{40})$ $\to A<(\dfrac1{10}+\dfrac1{10}+…+\dfrac1{10})+(\dfrac1{20}+\dfrac1{20}+…+\dfrac1{20})$ $\to A<1+1$ $\to A<2$ Vậy $1<A<2$ Trả lời
Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{11}+\dfrac1{12}+\dfrac1{13}+…+\dfrac1{40}$
$\to A=(\dfrac1{11}+\dfrac1{12}+…+\dfrac1{20})+(\dfrac1{21}+\dfrac1{22}+…+\dfrac1{40})$
$\to A>(\dfrac1{20}+\dfrac1{20}+…+\dfrac1{20})+(\dfrac1{40}+\dfrac1{40}+…+\dfrac1{40})$
$\to A>\dfrac12+\dfrac12$
$\to A>1$
Ta có:
$A=\dfrac{1}{11}+\dfrac1{12}+\dfrac1{13}+…+\dfrac1{40}$
$\to A=(\dfrac1{11}+\dfrac1{12}+…+\dfrac1{20})+(\dfrac1{21}+\dfrac1{22}+…+\dfrac1{40})$
$\to A<(\dfrac1{10}+\dfrac1{10}+…+\dfrac1{10})+(\dfrac1{20}+\dfrac1{20}+…+\dfrac1{20})$
$\to A<1+1$
$\to A<2$
Vậy $1<A<2$