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A = 54/15.18 + 54/18.21 + .........+ 54/87.90
A = 54/3 . ( 1/15.18 + 1/18.21 + ........+ 1/87.90)
A = 54/3 . ( 1/15 - 1/18 + 1/18 -1/21 + ......+ 1/87 - 1/90)
A =54/3 . ( 1/15 -1/90)
A = 54/3 . 1/18 = 1
vậy A = 1
a) (42 x 43 + 43 x 57 + 43) - 360 : 4
= 43 x ( 42 x 57) + 43 - 360 :4
= 43 x 99 + 43 - 90
= 4257 + 43 - 90
= 4300 - 90
= 4210
b) (372 - 194 x 4) + (981 : 9 - 13)
= (372 - 776) + (109 - 13)
= -404 + 96
= -308
c) 456 : 2 x 18 + 456 : 3 - 102
= 228 x 18 + 152 - 102
= 4104 + 152 - 102
= 4256 - 102
= 4154
=
a) (42.43+43.57+43)-360:4
=4300 - (360:4)
=4300- 90
=4210
b) (372 -194 .4 ) + ( 981 : 9 - 13 )
=712 + 69
=808
c) 456 : 2 . 18 + 456 : 3 - 102
= (456 : 2 . 18) + (456 : 3 - 102)
=4104+50
=4155
k nhak
a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0+3/4.x = 9/10
3/4.x = 9/10
x = 9/10: 3/4
x = 6/5
b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5
x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5
x + 3/2. ( 1-1/15) = 11/5
x + 3/2.14/15 = 11/5
x + 7/5 = 11/5
x = 11/5 - 7/5
x = 4/5
Tính :
a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\frac{2}{75}\)
\(=\frac{1}{75}\)
a; 5\(\dfrac{3}{4}\) : 3 + 2\(\dfrac{1}{4}\).\(\dfrac{1}{3}\) - \(\dfrac{3}{8}\)
= \(\dfrac{23}{4}\) : 3 + \(\dfrac{9}{4}\).\(\dfrac{1}{3}\) - \(\dfrac{3}{8}\)
= \(\dfrac{23}{4}\) x \(\dfrac{1}{3}\) + \(\dfrac{3}{4}\) - \(\dfrac{3}{8}\)
= \(\dfrac{23}{12}\) + \(\dfrac{3}{4}\) - \(\dfrac{3}{8}\)
= \(\dfrac{46}{24}\) + \(\dfrac{18}{24}\) - \(\dfrac{9}{24}\)
= \(\dfrac{64}{24}\) - \(\dfrac{9}{24}\)
= \(\dfrac{55}{24}\)
\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)
\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)
\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)
\(A=\frac{1}{4}-\frac{1}{104}\)
\(A=\frac{25}{104}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)
\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)
\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)
\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)
\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)
\(B\cdot2=\frac{2}{75}\)
\(B=\frac{2}{75}\frac{\cdot}{\cdot}2\)
\(B=\frac{1}{75}\)