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a) 3 x X x 2/3 = 4/5
3.x=6/5
x=2/5
b) 3,94 x 18,24 + 18,24 x 3,72 + 18,24 x 2,34
=18,24(3,94+3,72+2,34)
=18,24.10
=182,4
d) 8/19 x 2/7 + 8/19 x 1/7 + 8/19 x 4/7
=8/19(2/7+1/7+4/7)
=8/19.1
=8/19
\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999+2004}\).
Có sai đề không vậy???
Sửa đề một chút :v
\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999\cdot2004}\)
\(=\frac{1}{5}\left[\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{1999\cdot2004}\right]\)
\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{1999}-\frac{1}{2004}\right]\)
\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{2004}\right]\)
\(=\frac{1}{5}\cdot\frac{125}{501}=\frac{25}{501}\)
Đặt A =1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004. Ta có:
A= 1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004
5A= 5/4 x 9 + 5/9 x 14 + 5/14 x 19 +...+ 5/1999 + 2004
5A= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/1999 - 1/2004
5A= 1/4 - 1/2004
A= (1/4 - 1/2004)/5
\(a.=\frac{4}{13}\times\frac{13}{4}\times\frac{2}{3}\times\frac{3}{2}\)
\(=1\times1\)
\(=1\)
\(b.=\frac{19}{13}+\frac{7}{13}+\frac{14}{6}+\frac{4}{6}+\frac{1}{9}+\frac{17}{9}\)
\(=\frac{26}{13}+\frac{18}{6}+\frac{18}{9}\)
\(=2+3+2\)
\(=7\)
4/13 * 2/3 * 13/4* 3/2= (4/13 * 13/4 )* (2/3 * 3*2) = 1*1=1
19/13 + 14/6 + 1/9 + 4/6 + 7/13 + 17/9 = (19/13 + 7/13) + (14/6 + 4/6) + (1/9 + 17/9) = 2 + 3 + 2 =7
( 19 . x + 2 . 25 ) : 14 =9
(19.x+50):14=9
19.x+50=9.14
19.x+50=126
19.x=126-50
19.x=76
x=76:19
x=4
( 19 . x + 2 . 25 ) : 14 = 9
19 . x + 50 = 9 . 14
19 . x + 50 = 126
19x = 126 - 50
19x = 76
x = 76 : 19
x = 4