1/chứng minh rằng: 3^2 / 20*23+3^2 / 23*26+…+3^2 /77*80 nhỏ hơn 1 2/tìm x: 2/2*4+2/4*6+…+2/x*(x+2)=4/9 04/11/2021 Bởi Reagan 1/chứng minh rằng: 3^2 / 20*23+3^2 / 23*26+…+3^2 /77*80 nhỏ hơn 1 2/tìm x: 2/2*4+2/4*6+…+2/x*(x+2)=4/9
Đáp án: 2.$x=16$ Giải thích các bước giải: 1.Ta có :$A=\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+…+\dfrac{3^2}{77.80}$ $\to A=3(\dfrac{3}{20.23}+\dfrac{3}{23.26}+…+\dfrac{3}{77.80})$ $\to A=3(\dfrac{23-20}{20.23}+\dfrac{26-23}{23.26}+…+\dfrac{80-77}{77.80})$ $\to A=3(\dfrac1{20}-\dfrac1{23}+\dfrac1{23}-\dfrac1{26}+…+\dfrac1{77}-\dfrac1{80})$ $\to A=3(\dfrac1{20}-\dfrac{1}{80})$ $\to A<3\cdot\dfrac1{20}$ $\to A<3\cdot\dfrac13$ $\to A<1$ 2.Ta có :$\dfrac{2}{2.4}+\dfrac{2}{4.6}+…+\dfrac{2}{x(x+2)}=\dfrac49$ $\to\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+…+\dfrac{x+2-x}{x(x+2)}=\dfrac49$ $\to \dfrac12-\dfrac14+\dfrac14-\dfrac16+…+\dfrac1{x}-\dfrac{1}{x+2}=\dfrac49$ $\to\dfrac12-\dfrac1{x+2}=\dfrac49$ $\to \dfrac1{x+2}=\dfrac12-\dfrac49$ $\to\dfrac1{x+2}=\dfrac1{18}$ $\to x+2=18$ $\to x=16$ Bình luận
Đáp án: 2.$x=16$
Giải thích các bước giải:
1.Ta có :
$A=\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+…+\dfrac{3^2}{77.80}$
$\to A=3(\dfrac{3}{20.23}+\dfrac{3}{23.26}+…+\dfrac{3}{77.80})$
$\to A=3(\dfrac{23-20}{20.23}+\dfrac{26-23}{23.26}+…+\dfrac{80-77}{77.80})$
$\to A=3(\dfrac1{20}-\dfrac1{23}+\dfrac1{23}-\dfrac1{26}+…+\dfrac1{77}-\dfrac1{80})$
$\to A=3(\dfrac1{20}-\dfrac{1}{80})$
$\to A<3\cdot\dfrac1{20}$
$\to A<3\cdot\dfrac13$
$\to A<1$
2.Ta có :
$\dfrac{2}{2.4}+\dfrac{2}{4.6}+…+\dfrac{2}{x(x+2)}=\dfrac49$
$\to\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+…+\dfrac{x+2-x}{x(x+2)}=\dfrac49$
$\to \dfrac12-\dfrac14+\dfrac14-\dfrac16+…+\dfrac1{x}-\dfrac{1}{x+2}=\dfrac49$
$\to\dfrac12-\dfrac1{x+2}=\dfrac49$
$\to \dfrac1{x+2}=\dfrac12-\dfrac49$
$\to\dfrac1{x+2}=\dfrac1{18}$
$\to x+2=18$
$\to x=16$