(13/2020+23/2021-33/2022)-(1/2-1/3-1/6) tính nhanh
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\(\left(\dfrac{13}{2020}+\dfrac{23}{2021}+\dfrac{33}{2022}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{13}{2020}+\dfrac{23}{2021}+\dfrac{33}{2022}\right).\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{13}{2020}+\dfrac{23}{2021}+\dfrac{33}{2022}\right).0\)
=\(0\)
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
-13.21-13.80+13
=-13.(21-80+1)
=-13.100
=-1300
-1+2-3+4-5+6-...-2021+2022
=(-1+2)+(-3+4)+...+(-2021+2022)
=1+1+1+...+1 (1011 số hạng)
=1011
-13 . 21 - 13 . 80 + 13
13 ( -21 - 80 + 1 )
13 . ( -100 )
- 1300
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
= 1.495676488x10-3