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=(3/10/99+4/11/119-5/8/299)*(3/6-2/6-1/6)
=(3/10/99+4/11/119-5/8/299)*(1/6-1/6)
=(3/10/99+4/11/119-5/8/299)*0
=0
\(B=1+2+4+5+7+8+10+...+119+121+122\)
Ta có :
\(A=1+2+3+...+121+122\)
\(A=\left[\left(122-1\right):1+1\right]\left(1+122\right):2\)
\(A=122.\left(123\right):2=7503\)
Ta lại có :
\(C=3+6+9+...+120\)
\(C=\left[\left(120-3\right):3+1\right]\left(3+120\right):2\)
\(C=40.123:2=2460\)
Ta thấy : \(B=A-C\)
\(B=7503-2460=5043\)
a) \(\frac{1}{8}+\frac{3}{5}=\frac{5+24}{40}=\frac{29}{40}\)
b) \(\frac{6}{35}.\frac{49}{54}+\frac{1}{3}=\frac{6.7.7}{5.7.6.9}+\frac{1}{3}=\frac{7}{45}+\frac{1}{3}=\frac{7}{45}+\frac{15}{45}=\frac{22}{45}\)
c) \(\frac{4}{5}:\frac{3}{4}+0,25=\frac{4}{5}.\frac{4}{3}+\frac{25}{100}=\frac{16}{15}+\frac{1}{4}=\frac{64+15}{60}=\frac{79}{60}\)
Đặt A=1/1*3+1/3*5+...+1/119*121
2A=2/1*3+2/3.5+...+2/119.121
2A=1-1/3+1/3-1/5+...+1/119-1/121
2A=1-1/121
2A=120/121
A=60/121
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{119.121}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{119.12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{119}-\frac{1}{121}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{121}\right)\)
\(=\frac{1}{2}.\frac{120}{121}\)
\(=\frac{60}{121}\)