=\(11\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99\cdot100}\right)\)=\(11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)=\(11\left(1-\frac{1}{100}\right)\)=11\(\frac{99}{100}\)=\(\frac{1089}{100}\)
Đặt \(A=\frac{11}{1.2}+\frac{11}{2.3}+...+\frac{11}{99.100}\)
\(\Rightarrow A=11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=11\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=11.\frac{99}{100}\)
\(\Rightarrow A=\frac{1089}{100}\)
11/1.2+11/2.3+.....+11/99.100
.=11.(1/1.2+1/2.3+....+1/99+100)
=11.(1-1/2+1/2-1/3+.....+1/99-1/100)
=11.(1-1/100)
=11.99/100
=1089/100
gọi A bằng biể thức trên, ta có:
A=\(11.\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
A=\(11.\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+.....+\frac{100-99}{99.100}\right)\)
A=\(11.\left(\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+.....+\frac{100}{99.100}-\frac{99}{99.100}\right)\)
A=\(11.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
A=\(11.\left(1-\frac{1}{100}\right)\)
A=11.\(\frac{99}{100}\)
A=\(\frac{1089}{100}\)
Vậy A=... lười viết :V
