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a) 1+2+3+...+2018
=[(2018-1):1+1].(2018+1):2
=2018.2019:2
=2037171
b) 4+7+10+13+...+2017
=[(2017-4):3+1].(2017+4):2
=672.2021:2
=679056
c) 1+2-3-4+5+6-7-8+9+10-11-12+13+14
=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+13+14
=-4+(-4)+(-4)+13+14
=-12+27
=15
d) 23 x 75 + 25 x 23 + 180
=23 x (75+25) +180
=23 x 100 +180
=2300+180
2480
chúc ban hoc tốt nha
a) Số số hạng: ( 2018 - 1 ) : 1 + 1 = 2018
Tổng: 2018 x ( 2018 + 1 ) : 2 = 2 037 171
\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\times\left(1-\dfrac{1}{6}\right)\times\dots\times\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\) (sửa đề)
\(=\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dots\times\dfrac{98}{99}\times\dfrac{99}{100}\)
\(=\dfrac{2\times3\times4\times5\times\dots\times98\times99}{3\times4\times5\times6\times\dots\times99\times100}\)
\(=\dfrac{2}{100}\)
\(=\dfrac{1}{50}\)
a) x= \(\frac{-5}{12}\)
b) x = \(\frac{2}{5}\)
c) x =\(\frac{-87}{140}\)
d) x = \(\frac{109}{140}\)
e) x = \(\frac{13}{63}\)
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\(4,7\div0,25+5,3\times4\)
\(=18,8+21,2\)
\(=40\)
\(3\times\left(a-2\right)+150=240\)
\(3\times\left(a-2\right)=90\)
\(a-2=30\)
\(a=32\)
\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)
\(\dfrac{1}{9}+a=\dfrac{13}{36}\)
\(a=\dfrac{1}{4}\)
\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)
\(a=\dfrac{3}{2}\)
\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)
Rút gọn các thừa số ở tử và mẫu ta được:
\(B=\frac{1}{2004}\)
Đ/S:\(\frac{1}{2004}\)
Ta có:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)
Đơn giản hết sẽ là:
\(=\frac{1}{2004}\)
\(\frac{1}{2}\times\frac{1}{3}\times.........\times\frac{1}{10}=\frac{1}{2\times3}+......+\frac{1}{9\times10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(x:\frac{8}{9}=1\frac{3}{7}\)
\(x:\frac{8}{9}=\frac{10}{7}\)
\(x=\frac{10}{7}.\frac{8}{9}\)
\(x=\frac{80}{63}\)
\(x-1\frac{1}{4}=2\frac{1}{8}\)
\(x-\frac{5}{4}=\frac{17}{8}\)
\(x=\frac{17}{8}+\frac{5}{4}\)
\(x=\frac{17}{8}+\frac{10}{8}\)
\(x=\frac{27}{8}\)
\(x:\frac{8}{9}=1\frac{3}{7}\)
\(x:\frac{8}{9}=\frac{10}{7}\)
\(x=\frac{10}{7}\times\frac{8}{9}\)
\(x=\frac{80}{63}\)
#Y/n
1. \(1\frac{1}{5}x1\frac{1}{6}x1\frac{1}{7}x...x1\frac{1}{2006}\)
\(=\frac{6}{5}x\frac{7}{6}x\frac{8}{7}x...x\frac{2007}{2006}\)
\(=\frac{6x7x8x...x2007}{5x6x7x...x2006}\)
\(=\frac{2007}{5}.\)
2. Câu này đề chắc chắn sai.
3.\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{2600}\right)\)
\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2599}{2600}\)
\(=\frac{1x2x3x...x2599}{2x3x4x...x2600}\)
\(=\frac{1}{2006}.\)