ra số lẻ

9

800 triệu đồng chiếm:

\(1-\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{5}{12}\)(tổng số tiền)

Tổng số tiền là:

\(800000000:\dfrac{5}{12}=1920000000\left(đồng\right)\)

\(B=\dfrac{1}{3}\cdot b-\dfrac{2}{9}\cdot b-b:\dfrac{9}{4}\)

\(=\dfrac{1}{3}\cdot b-\dfrac{2}{9}\cdot b-\dfrac{4}{9}\cdot b\)

\(=b\left(\dfrac{1}{3}-\dfrac{2}{9}-\dfrac{4}{9}\right)\)

\(=b\left(\dfrac{1}{3}-\dfrac{6}{9}\right)=b\cdot\left(\dfrac{1}{3}-\dfrac{2}{3}\right)=-\dfrac{1}{3}b\)

Thay b=9/10 vào B, ta được:

\(B=\dfrac{-1}{3}\cdot\dfrac{9}{10}=-\dfrac{3}{10}\)

Đặt \(A=\dfrac{10^{2022}+1}{10^{2023}+1};B=\dfrac{10^{2021}+1}{10^{2022}+1}\)

\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)

Vì \(10^{2023}+1>10^{2022}+1\)

nên \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)

=>\(\dfrac{9}{10^{2023}+1}+1< \dfrac{9}{10^{2022}+1}+1\)

=>10A<10B

=>A<B

\(\dfrac{5}{x+1}=\dfrac{20}{-12}\\ \Rightarrow\dfrac{5}{x+1}=\dfrac{5}{-3}\\ \Rightarrow x+1=-3\\ \Rightarrow x=-4\)

\(\dfrac{5}{x+1}=\dfrac{20}{-12}\)

\(\Rightarrow\dfrac{5}{x+1}=\dfrac{5}{-3}\)

\(\Rightarrow x+1=-3\)

\(\Leftrightarrow x=-4\)

8,1 + (-3,7) + 1,9 + (-6,3)

=[(-3,7) + (-6,3)] + (8,1 + 1,9)

=10 - 10

=0

\(\text{8,1 + (-3,7) + 1,9 + (-6,3) }\)

\(\text{=[(-3,7) + (-6,3)] + (8,1 + 1,9) }\)

\(=\left(-10\right)+10\)

\(=0\)

(-100)+78=-(100-78)=-22

(-100)+78=-(100-78)=-22