Question

1.Tính nhanh 

a)427-98

b)2*19*15+3*43*10+62*80

c)\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

d)\(\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{2}{9}\right)\cdot\left(1-\frac{3}{90}\right)\cdot.........\cdot\left(1-\frac{2018}{9}\right)\)

Answer 1

a,427-98

=(427+2)-(98+2)

=429-100

=329

Answer 2

\(a)\) \(427-98=329\)

\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)

\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)

\(=30\cdot19+30\cdot43+62\cdot80\)

\(=30\cdot\left(19+43\right)+62\cdot80\)

\(=30\cdot62+62\cdot80\)

\(=62\cdot\left(30+80\right)\)

\(=62\cdot110=6820\)

\(c)\)  Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)

\(\Rightarrow2M=1-\frac{1}{3^6}\)

\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)

Vậy \(M=\frac{364}{729}\)