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a) \(12.79-64.37+12.21-36.37+12000\)
\(=\left(12.79+12.21\right)-\left(64.37+36.37\right)+12000\)
\(=\left[12\left(79+21\right)-37\left(64+36\right)+12000\right]\)
\(=12.100-37.100+12000\)
\(=1200-3700+12000\)
\(=9500\)
b) \(112.39+1237+112.61\)
\(=\left(112.39+112.61\right)+1237\)
\(=112\left(39+61\right)+1237\)
\(=112.100+1237\)
\(=11200+1237\)
\(=12437\)
c) \(35.762+351-35.662\)
\(=35\left(762-662\right)+351\)
\(=35.100+351\)
\(=3500+351\)
\(=3851\)
d) \(\frac{20}{4}-\frac{5^9}{5^8}=5-5=0\)
e) \(\left(\frac{5^{19}}{5^{17}}-4\right):7=\left(5^2-4\right):7=21:7=3\)
f) \(\frac{84}{4}+\frac{3^9}{3^7}+5^0=21+3^2+1=31\)
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
mk chỉ pik câu C thui
C= (1 - 1/2).(1 - 1/3).(1 - 1/4).......(1 - 4/4)....(1-1/2007)
C= (1 - 1/2).(1 - 1/3)..........0........(1 - 1/2007)
C= 0
cậu nên rút gọn bớt phần dài thay vào dó bằng ba dấu chấm . cậu cứ yên tâm đi thầy giáo chủ nhiệm lớp tớ bảo lảm kiểu như thế này ko sao đâu . cũng ko bị trừ điểm nữa
Bài cho ít số hạng không cần thiết phải viết ....