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\(2022\times2005-2000\times2022+15\times2022-20\times2021\)
\(=2022\times\left(2005-2000+15\right)-20\times2021\)
\(=2022\times20-20\times2021\)
\(=20\times\left(2022-2021\right)\)
\(=20\times1\)
\(=20\)
a, 2022 \(\times\) 2005 - 2000 \(\times\) 2022 + 15 \(\times\) 2022 - 20 \(\times\) 2021
= (2022 \(\times\) 2005 - 2000 \(\times\) 2022 + 15 \(\times\) 2022 )- 20 \(\times\) 2021
= 2022 \(\times\) (2005 - 2000 + 15) - 20 \(\times\) 2021
= 2022 \(\times\) (5 +15) - 20 \(\times\) 2021
= 2022 \(\times\) 20 - 20 \(\times\) 2021
= 20 \(\times\) (2022 - 2021)
= 20 \(\times\) 1
= 20
\(\dfrac{2021}{2022}\) x \(\dfrac{2022020222022}{202320232023}\) x \(\dfrac{20212021}{20232023}\)
= \(\dfrac{2021}{2022}\) x \(\dfrac{2022}{2023}\) x \(\dfrac{2021}{2023}\)
= \(\dfrac{2021\times2021}{2023\times2023}\)
= \(\dfrac{4084441}{4092529}\)
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
Nhỏ hơn
Ta có 2020/2021 <1
2021/2022 <1
2022/2023 <1
2023/2024 <1
Suy ra A=(2021/2021+2021/2022 +2022/2023 +2023/2024) < (1+1+1+1)= 4
Vậy A <4
Chúc bạn học tốt
\(\dfrac{2020}{2021}< 1\)
\(\dfrac{2021}{2022}< 1\)
\(\dfrac{2021}{2022}< 1\)
\(\dfrac{2023}{2024}< 1\)
Do đó: A<4
a. ( 48 x 0,75 - 240 : 10 ) - ( 16 x 0,5 - 16 : 4 )
=(36-24)-(8-4)
=12-4
=8
b. (1000 x 0,1 - 900 x 0,1 - 90 x 0,1 ) x ( 142 x 0,5 - 240 x 0,25 )
=(100-90)x(71-60)
=10x11
=110
# mui #
A = ( 2022 x 0,75 + 2022 : 4) - ( 2021 : 0,1 : 10)
= (2022 x 0,75 + 2022 x 0,25) - ( 2021 : 1/10 : 10)
= 2022 x(0,75+0,25) - 2021x 10 : 10
= 2022x 1 - 2021
= 2022 - 2021
=1