chứng minh rằng: 3^2 / 20*23+3^2 / 23*26+…+3^2 /77*80 nhỏ hơn 1 04/11/2021 Bởi aihong chứng minh rằng: 3^2 / 20*23+3^2 / 23*26+…+3^2 /77*80 nhỏ hơn 1
Giải thích các bước giải: Ta có :$A=\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+…+\dfrac{3^2}{77.80}$ $\to A=3(\dfrac{3}{20.23}+\dfrac{3}{23.26}+…+\dfrac{3}{77.80})$ $\to A=3(\dfrac{23-20}{20.23}+\dfrac{26-23}{23.26}+…+\dfrac{80-77}{77.80})$ $\to A=3(\dfrac1{20}-\dfrac1{23}+\dfrac1{23}-\dfrac1{26}+…+\dfrac1{77}-\dfrac1{80})$ $\to A=3(\dfrac1{20}-\dfrac{1}{80})$ $\to A<3\cdot\dfrac1{20}$ $\to A<3\cdot\dfrac13$ $\to A<1$ Bình luận
Giải thích các bước giải:
Ta có :
$A=\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+…+\dfrac{3^2}{77.80}$
$\to A=3(\dfrac{3}{20.23}+\dfrac{3}{23.26}+…+\dfrac{3}{77.80})$
$\to A=3(\dfrac{23-20}{20.23}+\dfrac{26-23}{23.26}+…+\dfrac{80-77}{77.80})$
$\to A=3(\dfrac1{20}-\dfrac1{23}+\dfrac1{23}-\dfrac1{26}+…+\dfrac1{77}-\dfrac1{80})$
$\to A=3(\dfrac1{20}-\dfrac{1}{80})$
$\to A<3\cdot\dfrac1{20}$
$\to A<3\cdot\dfrac13$
$\to A<1$