Bài giải

4 năm trước, tổng số tuổi của bố và con là:

68-4-4=60(tuổi)

Tổng số phần bằng nhau là:

3+1=4(phần)

Tuổi của bố 4 năm trước là:

60:4x3=45(tuổi)

Tuổi của bố hiện nay là:

45+4=49(tuổi)

Tuổi của bố 4 năm sau là:

49+4=53(tuổi)

Tuổi của con hiện nay là:

68-49=19(tuổi)

Tuổi của con 4 năm sau là:

19+4=23(tuổi)

Đáp số: Bố: 53 tuổi

Con: 23 tuổi

\(\dfrac{4}{9}-\dfrac{1}{6}+\dfrac{5}{9}+2\\ =\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+2-\dfrac{1}{6}\\ =\dfrac{9}{9}+2-\dfrac{1}{6}\\ =1+2-\dfrac{1}{6}\\ =\left(1+2\right)-\dfrac{1}{6}\\ =3-\dfrac{1}{6}\\ =\dfrac{18}{6}-\dfrac{1}{6}=\dfrac{17}{6}\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}=\\ A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{99.99}+\dfrac{1}{100.100}\\ A< \dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\\ A< \dfrac{1}{2.2}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\\ A< \dfrac{1}{2.2}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ A< \dfrac{1}{2.2}+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ A< \dfrac{1}{2.2}+\dfrac{49}{50}\\ A< \dfrac{1}{4}+\dfrac{49}{50}\\ A< \dfrac{37}{50}=\dfrac{74}{100}< \dfrac{75}{100}=\dfrac{3}{4}\) Hay \(A< \dfrac{3}{4}\)

Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{3}{4}\)

\(\dfrac{8}{5}:\left(2\times\dfrac{4}{3}-\dfrac{5}{6}\right)\\ =\dfrac{8}{5}:\left(\dfrac{8}{3}-\dfrac{5}{6}\right)\\ =\dfrac{8}{5}:\left(\dfrac{16}{6}-\dfrac{5}{6}\right)\\ =\dfrac{8}{5}:\dfrac{11}{6}\\ =\dfrac{8}{5}\times\dfrac{6}{11}\\ =\dfrac{48}{55}\)

\(32,1+5,1\times32,1+5,9-32,1\\ =32,1\times\left(1+5,1-1\right)+5,9\\ =32,1\times5,1+5,9\\ =163,71+5,9\\ =169,61\)

Ta có công thức: \(\dfrac{n\left(n-1\right)}{2}\)

Thay vào bài, ta được:

\(\dfrac{n\left(n-1\right)}{2}=91\\ n\left(n-1\right)=91.2\\ n\left(n-1\right)=182\\ 14\left(14-1\right)=182\)

Vậy \(n=14\)

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\)

Ta có: \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}-\dfrac{1}{100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{49}{100}< \dfrac{50}{100}=\dfrac{1}{2}\)

Hay \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}\) 

Vì \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

Vậy biểu thức \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{90}\\ =\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{90}\\ =\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}\right)+\dfrac{1}{90}\\ =\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)+\dfrac{1}{90}\\ =\left(1-\dfrac{1}{5}\right)+\dfrac{1}{90}\\ =\left(\dfrac{5}{5}-\dfrac{1}{5}\right)+\dfrac{1}{90}\\ =\dfrac{4}{5}+\dfrac{1}{90}\\ =\dfrac{72}{90}+\dfrac{1}{90}=\dfrac{73}{90}\)

\(\dfrac{3}{4}+\dfrac{3}{4}\\ =\dfrac{3+3}{4}\\ =\dfrac{6}{4}\\ =\dfrac{6:2}{4:2}\\ =\dfrac{3}{2}\)

\(\dfrac{2021\times2023-1}{2020\times2023+2022}\\ =\dfrac{2023\times\left(2020+1\right)-1}{2023\times2020+2022}\\ =\dfrac{2023\times2020+2023\times1-1}{2023\times2020+2022}\\ =\dfrac{2023\times2020+2023-1}{2023\times2020+2022}\\ =\dfrac{2023\times2020+\left(2023-1\right)}{2023\times2020+2022}\\ =\dfrac{2023\times2020+2022}{2023\times2020+2022}\\ =1\)