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(1+3+3+7+...+97+99)x(45x3-45x2-45)
Ta xét tổng:45x3-45x2-45
=45x3-45x2-45x1
=45x (3-2-1)
=45x0
=0
Thay vào biểu thức trên ta có:(1+3+3+7+...+97+99)x0
=0
Vậy gía trị biểu thức trên là 0
\(\frac{9}{10}\)+\(\frac{7}{9}\)+\(\frac{5}{8}\)+\(\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)\)\(+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=1+1+1+1+1\)
\(=5\)
9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
= ( 9/10 + 1/10 ) + ( 7/9 + 2/9 ) + ( 5/8 + 3/8 ) + ( 3/7 + 4/7 ) + ( 3/5 + 2/5 )
= 1 + 1 + 1 + 1 + 1
= 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)
= \(1+1+1+1+1\)
= \(1\times5\)
= \(5\)
Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :
A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)
A = 1 + 1 + 1 + 1 + 1
A = 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)\(=\frac{9}{10}+\frac{1}{10}+\frac{7}{9}+\frac{2}{9}+\frac{5}{8}+\frac{3}{8}+\frac{3}{7}+\frac{4}{7}+\frac{3}{5}+\frac{2}{5}\)
\(=\left[\frac{9}{10}+\frac{1}{10}\right]+\left[\frac{7}{9}+\frac{2}{9}\right]+\left[\frac{5}{8}+\frac{3}{8}\right]+\left[\frac{3}{7}+\frac{4}{7}\right]+\left[\frac{3}{5}+\frac{2}{5}\right]\)
\(=1+1+1+1+1\)
\(=5\)
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= 1 + 1 + 1 + 1 + 1
= 5
\(\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{3}{5}+\frac{1}{6}\)
\(=\frac{3}{6}-\frac{2}{5}+\frac{2}{6}-\frac{3}{5}+\frac{1}{6}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\)
\(=1+1\)
\(=2\)
@muối
1/2 - 2/5 + 1/3 - 3/5 + 1/6
= 1/2 + 1/3 + 1/6 - 2/5 - 3/5
= (1/2 + 1/3 + 1/6) - (2/5 + 3/5)
= 1 - 1
= 0
Chúc bạn học tốt.
😁😁😁
1+2+3+4+5+6+7+8+9
= (9+1)+(8+2)+(7+3)+(6+4)+5
= 10 + 10 + 10 + 10 + 5
= 40 + 5
= 45
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{101-99}{99x101}=\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)