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Bài làm
\(A=\frac{0,24\cdot450+0,8\cdot15\cdot5+3\cdot3\cdot8}{65-60+55-50+45-40+35-30+25-20+5}\)
\(A=\frac{0,24\cdot45\cdot10+0,8\cdot15\cdot5+3\cdot3\cdot8}{\left(65-60\right)+\left(55-50\right)+\left(45-40\right)+\left(35-30\right)+\left(25-20\right)+5}\)
\(A=\frac{2,4\cdot45+0,8\cdot75+72}{5+5+5+5+5+5}\)
\(A=\frac{2,4\cdot45+0,8\cdot\frac{5}{3}\cdot45+1,6\cdot45}{30}\)
\(A=\frac{45\left(2,4+\frac{4}{5}\cdot\frac{5}{3}+1,6\right)}{30}\)
\(A=\frac{45\cdot\left(\frac{20}{15}+\frac{36}{15}+\frac{24}{15}\right)}{30}\)
\(A=\frac{3\cdot\frac{80}{15}}{2}\)
\(A=\frac{\frac{80}{5}}{2}\)
\(A=\frac{80}{5}\cdot\frac{1}{2}\)
\(A=\frac{40}{5}\)
\(A=8\)
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5/6 + 11/12 + 19/20 + ... + 109/110
= (1 - 1/6) + (1 - 1/12) + (1 - 1/20) + ... + (1 - 1/110)
= (1 - 1/2×3) + (1 - 1/3×4) + (1 - 1/4×5) + ... + (1 - 1/10×11)
= (1 + 1 + 1 + ... + 1) - (1/2×3 + 1/3×4 + 1/4×5 + ... + 1/10×11)
( có 9 số 1)
= 1 x 9 - (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/10 - 1/11)
= 9 - (1/2 - 1/11)
= 9 - (11/22 - 2/22)
= 9 - 9/22
= 198/22 - 9/22
= 189/22
☆☆☆☆☆
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a)135-35.(-37)+37.(-42-48)=-4490
b)-65.(87-17)-87.(17-65)=-374
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11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110 =
256 phần 33
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11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110
= \(\frac{256}{53}\)
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a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
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