Bài 1 :
Đặt A=1.2+2.3+3.4+4.5+.........+99.100
=> 3A=1.2.3+2.3.(4-1)+........+99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+........+99.100.101-98.99.100
3A=99.100.101
A=33.100.101
A=333300
Bài 2 :
1:20 + 1:44 + 1:77 + 1:119 + 1:170 = \(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}=\frac{1}{10}=0,1\)
1)1.2+2.3+3.4+4.5+...+99.100
đặt 3D=1.2+2.3+3.4+...+99.100
=1.2.3+2.8.3+...+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5
=99.100.101
=999900
D=999900:3=333300
nếu đúng nhớ cảm ơn nhak. mình ko bít làm bài 2
\(\frac{1}{20}+\left(\frac{1}{44}+\frac{1}{77}\right)+\left(\frac{1}{119}+\frac{1}{170}\right)=\frac{1}{20}+\left(\frac{1}{11}.\frac{1}{4}+\frac{1}{11}.\frac{1}{7}\right)+\left(\frac{1}{17}.\frac{1}{7}+\frac{1}{17}.\frac{1}{10}\right)\)
= \(\frac{1}{20}+\frac{1}{11}.\left(\frac{1}{4}+\frac{1}{7}\right)+\frac{1}{17}.\left(\frac{1}{7}+\frac{1}{10}\right)=\frac{1}{20}+\frac{1}{11}.\frac{11}{28}+\frac{1}{17}.\frac{17}{70}=\frac{1}{20}+\frac{1}{28}+\frac{1}{70}\)
= \(\frac{1}{20}+\frac{1}{14}.\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{1}{20}+\frac{1}{14}.\frac{7}{10}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=0,1\)
