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A=100+98+96+...+2−97−95−...1A=100+98+96+...+2−97−95−...1
A=100+(98−97)+(96−95)+...(2−1)A=100+(98−97)+(96−95)+...(2−1)
A=100+1+1+1+...+1A=100+1+1+1+...+1
A=100+1.49A=100+1.49
A=100+49A=100+49
A=149
a, 100 + 98 + 96 + ... + 2 - 9 7 - 95 - .. -1
= 100 + (98 - 97) + (96-95) + ... + + ... + (2 - 1)
= 100 + 1 + 1 + 1 +.. +1
= 100 + 1 x 49
= 100 + 49
= 149
b , 1 + 2 - 3 - 4 + 5 + 6 - .... -299 - 330 +301 + 302
=( 1 + 2 - 3) + ( -4 + 5 + 6 -7 ) +... +(298 - 299 -300 +301 ) + 302
= 0 + 0 + .. + 0 + 302
= 302
a) 0,4(3) = \(\frac{4,\left(3\right)}{10}=\frac{4+\frac{1}{3}}{10}=\frac{13}{30}\); 0,6(2) = \(\frac{6,\left(2\right)}{10}=\frac{6+\frac{2}{9}}{10}=\frac{56}{90}=\frac{28}{45}\); 0,5(8) = \(\frac{5,\left(8\right)}{10}=\frac{5+\frac{8}{9}}{10}=\frac{53}{90}\)
Vậy A = \(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{53}{90}}:\frac{2700}{53}\) = \(\frac{13}{30}+\frac{14}{9}-\frac{5}{6}.\frac{90}{53}.\frac{53}{2700}=\frac{13}{30}+\frac{14}{9}-\frac{1}{36}=\frac{353}{180}\)
b) 0,(5) = 5/9; 0,(2) = 2/9
B = \(\left(\frac{5}{9}.\frac{2}{9}\right):\left(\frac{10}{3}.\frac{25}{33}\right)-\left(\frac{2}{5}.\frac{4}{3}\right):\frac{4}{3}\)
B = \(\frac{10}{81}.\frac{3.33}{10.25}-\frac{2}{5}=\frac{11}{225}-\frac{2}{5}=-\frac{79}{225}\)
A = 6 -2/3 + 1 /2 - 5 -5/3 +3/2 -3 + 7/3 - 5/2
= (6 - 5 - 3) - ( 2/3 -5/3 + 7/3 ) + ( 1/2 +3/2 - 5/2)
= -2 + 0 -1/2 = -5/2