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a ) 1 + 2 + 3 + 4 + ... + x = 1275 ( có x số tự nhiên )
( x + 1 ) . x : 2 = 1275
( x + 1 ) . x = 1275 x 2
( x + 1 ) . x = 2550
( x + 1 ) . x = 50 . 51
Mà x , x + 1 là hai số tự nhiên liên tiếp => x = 50
Vậy x = 50
1+2+3+4+...+x=1275
\(\frac{x.\left(x+1\right)}{2}=1275\)
x(x+1)=1275x2=2550
x(x+1)=50.51
x=50
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Bài 1:
a) [ (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15) phần 1/2 - 1/3 + 1/4 - 1/5 ] : (1/4 - 1/6)
= [ (1/6 : 1/6) + (1/10 : 1/10) - (1/15 : 1/15) phần 30/60 - 20/60 + 15/60 - 12/60 ] : (3/12 - 2/12)
= [ 1 + 1 - 1 phần 13/60 ] : 1/12
= [ 1 : 13/60 ] x 12
= 60/13 x 12
=720/ 13
b) (3/20 + 1/2 - 1/15) x 12/49 phần 3 và 1/3 + 2/9
= (9/60 + 30/60 - 4/60) x 12/49 phần 10/3 + 2/9
= 7/12 x 12/49 phần 30/9 + 2/9
= 1/7 : 32/9
= 1/7 x 9/32
= 9/224
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B =1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
B = 1 - 1/3 + 1/3 - 1/6 + 1/6 - 1/10 + 1/10 - 1/15 + 1/15 - 1/21 + 1/21 - 1/28
B = 1 - ( 1/3 + 1/3 - 1/6 + 1/6 - 1/10 + 1/10 - 1/15 + 1/15 - 1/21 + 1/21 ) - 1/28
B = 1 - 1/28
B = 27/28
~ Hok T ~
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1/3+1/6+1/10+1/15+......+1/4950
=2x(1/6+1/12+1/20+1/30+……+1/9900)
=2x(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100)
=2x(1/2-1/100)
=1-1/50
=49/50
**** nhé
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\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
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Bài làm:
Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)
\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)
\(=\frac{1}{2}.\frac{11}{12}\)
\(=\frac{11}{24}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)
\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)
\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)
\(=2\times\left(1-\frac{1}{12}\right)\)
\(=2\times\frac{11}{12}\)
\(=\frac{11}{6}\)