Question

[Sakura Kinomoto]

BÀI TẬP: THỰC HIỆN PHÉP TÍNH;

a.\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{97.100}\)                           b.\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

c.\(\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{\frac{49}{11}-\frac{7}{8}-\frac{35}{12}}\)                                                d.\(\frac{5-\frac{5}{3}-\frac{5}{9}-\frac{5}{27}}{-8+\frac{8}{3}+\frac{8}{9}+\frac{8}{27}}\div\frac{15-\frac{15}{11}-\frac{15}{121}}{16-\frac{16}{11}-\frac{16}{121}}\)

e.\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)}\)                                         f.\(\frac{3}{5}-\frac{2}{3}\left(1-0,3\right)-\left(-3\right)^2\)  

Answer 1

[Thanh Tùng DZ]

a) đặt B = \(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{97.100}\)

\(B=5.\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)\)

đặt A = \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\)

A = \(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+...+\frac{1}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)\)

A = \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

A = \(\frac{1}{3}.\left(1-\frac{1}{100}\right)\)

A = \(\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

\(\Rightarrow B=5.\frac{33}{100}=\frac{33}{20}\)

b) đặt C = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

3C = \(3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

3C - C = \(\left(3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)

2C = \(3-\frac{1}{3^{100}}\)

\(\Rightarrow\)C = \(\frac{3-\frac{1}{3^{100}}}{2}\)

c) \(\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{\frac{49}{11}+\frac{7}{8}-\frac{35}{12}}=\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{7.\left(\frac{7}{11}+\frac{1}{8}-\frac{5}{12}\right)}=\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{7.\left(\frac{5}{12}+\frac{1}{8}-\frac{7}{11}\right)}=\frac{1}{7}\)

e) \(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)}=\frac{2^3.5^3.7^4}{2.5.7^2}=2^2.5^2.7^2=70^2\)

f) \(\frac{3}{5}-\frac{2}{3}\left(1-0,3\right)-\left(-3\right)^2\)

\(=\frac{3}{5}-\frac{2}{3}.\frac{7}{10}+9=\frac{3}{5}-\frac{7}{15}+9=\frac{2}{15}+9=\frac{137}{15}\)

Answer 2

[Sakura Kinomoto]

cám ơn bn nhé