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\(432432\cdot468-234234\cdot864\)
= \(202378176-202378176\)
\(=0\)
Ta có :
\(\frac{2017\times2018+1}{2019+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018\times\left(2016+1\right)}\)
\(=\frac{2017\times2018+1}{1+2018\times2017}\)
\(=1\)
\(\frac{2017.2018+1}{2019+2016.2018}\)
\(=\frac{2017.2018+1}{1+2018+2016.2018}\)
\(=\frac{2017.(2018+1)}{(1+2018).\left(2016+1\right)}\)
\(=\frac{2017.2019}{2019.2017}\)
\(=\frac{1}{1}=1\)
\(\frac{3}{4\times7}+\frac{1}{7\times8}+\frac{5}{8\times13}+\frac{2}{13\times15}+\frac{9}{15\times24}\)
= \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{24}\)
= \(\frac{1}{4}-\frac{1}{24}\)
= \(\frac{6}{24}-\frac{1}{24}\)
= \(\frac{5}{24}\)
\(\frac{3}{4.7}+\frac{1}{1.8}+\frac{5}{8.13}+\frac{2}{13.15}+\frac{9}{15.24}\)
Đặt A = ( 3 + 1 + 5 + 2 + 9 ) . \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{24}\)
A = 20 . \(\frac{1}{4}-\frac{1}{24}\)
A = 20 . \(\frac{6}{24}-\frac{1}{24}\)
A = 20 . \(\frac{5}{24}\)
A = \(\frac{100}{24}\)
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# DanLinh
d ( 1-1/2)x(1-1/3)x(1-1/4)x......x(1-1/2018)
= 1/2x2/3x3/4x...x2017/2018
=\(\frac{1x2x3x....x2017}{2x3x4x....x2018}\)
= \(\frac{1}{2018}\)
e , 1+4+7+...+100
= dãy có số số hạng là
(100-1):3+1=34 ( số số hạng)
tổng là : (100+1 ) x 34 : 2 =1717
=>1717
=4,7.(15,6-11,4)+4,2.5,3
=4,7.4,2+4,2.5,3
=4,2.(4,7+5,3)
=4,2.10
=42
a, A=50(100+2):2-49(97+1):2=149
b, B=(1-3)+(2-4)+(5-7)+(6-8)+...+(297-299)+(298-300)+301+302
=-2-2-2-...-2+301+302(150 chữ số 2)
=-2.150+603
=330
a) mk chỉnh đề
\(A=\left(1+\frac{1}{2005}\right)\left(1+\frac{1}{2006}\right)\left(1+\frac{1}{2019}\right)\)
\(=\frac{2006}{2005}.\frac{2007}{2006}.....\frac{2020}{2019}\)
\(=\frac{2020}{2005}\)
\(=\frac{404}{401}\)
\(B=\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+....+\frac{3}{1+2+3+...+100}\)
\(=3+3\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=3+3.\left(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{100.101}{2}}\right)\)
\(=3+3.\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{100.101}\right)\)
\(=3+6\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=3+6\left(\frac{1}{2}-\frac{1}{101}\right)=3+6.\frac{99}{202}\)
\(=3+2\frac{95}{101}=5\frac{95}{101}\)
\(\)
a)
(x*211)+(1+2+...+211)=23632
(x*211)+22366=23632
x*211=23632-22366
x*211=1266
x=1266:211
x=6
Vậy x=6
\(\frac{2015+2016.2017}{2017.2018-2019}\)
\(=\frac{2015+2016.2017}{2017.\left(2016+2\right)-2019}\)
\(=\frac{2015+2016.2017}{2017.2016+4034-2019}\)
\(=\frac{2015+2016.2017}{2017.2016+2015}\)
\(=1\)
câu b cái ngoặc 2 bằng 0