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bài này mik làm rồi
mik bảo đảm đo
tick mik thật nhiều nhé các bạn
A=1+2+3+4+5+...+99+100
A=(1+100).100:2=101.50=5050
B=1/2+1/6+1/12+1/20+1/30+...+1/9900
B=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+....+1/99.100
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100
B=1-1/100=99/100
A = 100 x 101 : 2 = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Đề là chứng minh điểu thức bằng 2 phải không bạn?
\(\frac{200-\left(3+\frac{2}{3}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
\(=\frac{2.100-3-\frac{2}{3}-...-\frac{2}{100}}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
\(=\frac{\left(4-3\right)+\left(2-\frac{2}{3}\right)+\left(2-\frac{2}{4}\right)+...+\left(2-\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
\(=\frac{1+\frac{4}{3}+\frac{6}{4}+...+\frac{198}{100}}{1+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
\(=2\)
\(\)
\(\frac{200-\left(3+\frac{2}{3}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)
=\(\frac{2x100-3-\frac{2}{3}-...-\frac{2}{100}}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
=\(\frac{\left(4-3\right)+\left(2-\frac{2}{3}\right)+\left(2-\frac{2}{4}\right)+...+\left(2-\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
=\(\frac{1+\frac{4}{3}+\frac{6}{4}+...+\frac{198}{100}}{1+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
=\(\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
=\(2\)
Ta có \(A=\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{99}{100}}\)
\(A=\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{100}\right)}{\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)}\)
\(A=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\)
\(\Rightarrow A=2\)