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\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)
\(\frac{3}{2}\cdot\frac{1}{3}+\frac{8}{5}:4\\ =\frac{1}{2}+\frac{8}{5\cdot4}\\ =\frac{1}{2}+\frac{2}{5}\\ =\frac{5}{10}+\frac{4}{10}\\ =\frac{9}{10}\)
3/2.1/3+8/5:4
=1/2+8/5.1/4
=1/2+2/5
=5/10+4/10
=9/10
NHỚ TICK MÌNH NHÉ
= 2/3 + 1/3 . 7/18 : 5/12
= 2/3 + 7/54 : 5/12
= 2/3 + 14/45
= 44/45
1-2+3-4+...+2021-2022+2023
=(1-2)+(3-4)+...+(2021-2022)+2023
=(-1)+(-1)+(-1)+...+(-1)+2023
=(-1011)+2023
=1012
Từ 2 đến 2011 có 2010 số hạng và 2012
= (2-3)+(4-5)+....+(2008-2009)+(2010-2011)+2012
= (-1)+(-1)+....+(-1)+(-1) +2012
Vì mỗi cặp 2 số => Có 1006 cặp
=(-1) x 1006 +2012
=1006+2012
=3018
Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{98^2}\)
Ta có : \(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{97.98}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{97}-\frac{1}{98}\)
\(A< \frac{1}{2}-\frac{1}{98}\)
\(A< \frac{1}{2}\)
GOOD LUCK !!!
Câu 1:
$B=\frac{10}{1.3}+\frac{10}{3.5}+\frac{10}{5.7}+...+\frac{10}{101.103}$
$B=5(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103})$
$=5(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{103-101}{101.103})$
$=5(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103})$
$=5(1-\frac{1}{103})=5.\frac{102}{103}=\frac{510}{103}$
Câu 2:
\(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2022.2024}\\ =\frac{1}{2}\left[\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2022.2024}\right]\)
\(=\frac{1}{2}\left[\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{2024-2022}{2022.2024}\right]\)
\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2022}-\frac{1}{2024})\\ =\frac{1}{2}(\frac{1}{2}-\frac{1}{2024})=\frac{1011}{4048}\)
1+2+3+4+...+2022
= (2022+1) . 2022 : 2
= 2023 . 1011
= 2 045 253
Cảm ơn bn nha :3