(3 * 20212021 * 20212023 -5 * 20212020 -2 * 20212021^2 -5)/20212021
giúp minh nhe cẻm un
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Đặt 20212020=x
=>\(A=\dfrac{3\left(x+1\right)\left(x+3\right)-5x-2\cdot\left(x+1\right)^2-5}{\left(x+1\right)}\)
\(=\dfrac{3\left(x^2+4x+3\right)-5x-2x^2-4x-2-5}{\left(x+1\right)}\)
\(=\dfrac{3x^2+12x+9-2x^2-9x-7}{x+1}=\dfrac{x^2+3x+2}{x+1}=x+2\)
=20212022
-2021 . 20222022 + 2022 .20212021
= -2021 . 2022. 10001 + 2022.2021. 10001
= 0
-2021 . 20222022 + 2022 . 20212021
= (-2021) . (20222022 - 2022 . 10001)
= (-2021) . (20222022 - 20222022)
= (-2021) . 0
= 0
\(\frac{20212021}{20222022}=\frac{20212021:10001}{20222022:10001}=\frac{2021}{2022}\)
\(\dfrac{2021}{2022}\) x \(\dfrac{2022020222022}{202320232023}\) x \(\dfrac{20212021}{20232023}\)
= \(\dfrac{2021}{2022}\) x \(\dfrac{2022}{2023}\) x \(\dfrac{2021}{2023}\)
= \(\dfrac{2021\times2021}{2023\times2023}\)
= \(\dfrac{4084441}{4092529}\)
\(\dfrac{121212}{242424}+\dfrac{1313}{2626}+\dfrac{20212021}{40424042}\)
= \(\dfrac{12}{24}+\dfrac{13}{26}+\dfrac{2021}{4042}\)
= \(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)
= \(\dfrac{3}{2}\)
121212/242424+1313/2626+20212021/404244024
= 1/2+1/2+1/2
=3/2