Tính hợp lí : 2022.2022 - 2025.2019
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a) Ta có:
\(A=2021\cdot2023\)
\(A=\left(2022-1\right)\cdot\left(2022+1\right)\)
\(A=2022^2+2022-2022-1\)
\(A=2022^2-1\)
Ta thấy: \(2022^2-1< 2022^2\)
Vậy: \(A< B\)
b) Ta có:
\(A=2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(B=3^{20}=\left(3^2\right)^{10}=9^{10}\)
Ta thấy: \(8^{10}< 9^{10}\)
Vậy: \(A< B\)
a)\(B=2022.2022\)
\(=\left(2021+1\right)2022\)
\(=2021.2022+2022\)
Vì \(2020< 2022\Rightarrow2021.2020< 2021.2022\Rightarrow2021.2020< 2021.2022+2022\)
hay \(A< B\)
b)\(M=360.345-200\)
\(=360\left(344+1\right)-200\)
\(=360.344+360-200\)
\(=360.344+160< 340.344+161=N\)
\(\Rightarrow M< N\)
a>b nhé anh vì 2022.2022 nó đã lớn hơn 2021.2023 rồi ạ
k cho em nếu đúng nhé
chúc anh học tốt
Ta có:
A=2022x2022
=2022x(2023-1)
=2022 x 2023 - 2022 x 1
=2022x2024-2021
Lại có:
B = 2021 . 2023
B = ( 2022 - 1 ) . 2023
B = 2023 . 2022 - 2023 . 1
B = 2023 . 2022 - 2023
Ta thấy: 2022x2024-2021 > 2023x2022-2023
Suy ra A > B.
\(\left(-2021\right).\left(23+76\right)-2021=\left(-2021\right).99-2021=-202100\)
\(5,-\dfrac{2}{7}-\left(\dfrac{5}{11}-\dfrac{9}{7}\right)+\dfrac{5}{11}=-\dfrac{2}{7}-\dfrac{5}{11}+\dfrac{9}{7}+\dfrac{5}{11}=\left(-\dfrac{2}{7}+\dfrac{9}{7}\right)-\left(\dfrac{5}{11}-\dfrac{5}{11}\right)=1-0=1\\ 6,\dfrac{2}{3}+\dfrac{-2}{5}+\dfrac{-5}{6}-\dfrac{13}{10}=\left(\dfrac{4}{6}+\dfrac{-5}{6}\right)+\left(\dfrac{-4}{10}-\dfrac{13}{10}\right)=\dfrac{-1}{6}+\dfrac{-17}{10}=\dfrac{-5}{30}+\dfrac{-51}{30}=\dfrac{-56}{30}=\dfrac{-28}{15}\)
\(7,\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{4}{-9}+\dfrac{7}{15}=\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)=\left(-1\right)+1=0\\ 8,\left(\dfrac{12}{13}+7\right)-\left(-\dfrac{1}{13}+8\right)=\dfrac{12}{13}+7+\dfrac{1}{13}-8=\left(\dfrac{12}{13}+\dfrac{1}{13}\right)+\left(7-8\right)=1+\left(-1\right)=0\)
a) \(\left(\dfrac{3}{8}-\dfrac{1}{5}-\dfrac{1}{3}\right)+\left(\dfrac{5}{8}-1\dfrac{3}{7}-\dfrac{4}{7}\right)+\dfrac{1}{3}\)
\(=\dfrac{3}{8}-\dfrac{1}{5}-\dfrac{1}{3}+\dfrac{5}{8}-\dfrac{10}{7}-\dfrac{4}{7}+\dfrac{1}{3}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-10}{7}-\dfrac{4}{7}\right)-\dfrac{1}{5}\)
\(=1+0+\left(-2\right)-\dfrac{1}{5}\)
\(=-1-\dfrac{1}{5}\)
\(=\dfrac{-6}{5}\)
b) \(-24.\left(\dfrac{1}{3}-\dfrac{1}{4}\right)^2-1,5\)
\(=-24.\left(\dfrac{1}{12}\right)^2-\dfrac{3}{2}\)
\(=-24.\dfrac{1}{144}-\dfrac{3}{2}\)
\(=\dfrac{-1}{6}-\dfrac{3}{2}=\dfrac{-1}{6}-\dfrac{9}{6}=\dfrac{-10}{6}=\dfrac{-5}{3}\)
c) \(2^2-\left(-\dfrac{5}{7}\right)^0+\left(\dfrac{1}{3}\right)^4.3^6\)
\(=4-1+\dfrac{1}{81}.729\)
\(=4-1+9\)
\(=3+9=12\)
d) \(23\dfrac{1}{3}:\left(\dfrac{-5}{7}\right)+13\dfrac{1}{3}:\dfrac{5}{7}\)
\(=\dfrac{-70}{3}.\dfrac{7}{5}+\dfrac{40}{3}.\dfrac{7}{5}\)
\(=\dfrac{7}{5}\left(\dfrac{-70}{3}-\dfrac{40}{3}\right)\)
\(=\dfrac{7}{5}.\left(-10\right)\)
\(=-14\)
e) \(\dfrac{3}{4}.\dfrac{8}{5}+\dfrac{6}{15}:\dfrac{4}{3}-1\dfrac{2}{5}:1\dfrac{1}{3}\)
\(=\dfrac{3}{4}.\dfrac{8}{5}+\dfrac{6}{15}.\dfrac{3}{4}-\dfrac{7}{5}.\dfrac{3}{4}\)
\(=\dfrac{3}{4}\left(\dfrac{8}{5}+\dfrac{6}{15}-\dfrac{7}{5}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{24}{15}+\dfrac{6}{15}-\dfrac{21}{15}\right)\)
\(=\dfrac{3}{4}.\dfrac{9}{15}\)
\(=\dfrac{9}{20}\)
25.8.4.28.125
= 28.(25.4).(8.125)
= 28.100.1000
= 2800.1000
= 2800000
2025-2022 =3 ×2019 = 6057