Tìm phân số thứ 50 ở dãy số: 3/2x7+3/7x12+3/12x17+3/17x22+3/22x27+3/27x32+...
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30 tháng 7 2021
Ta có:\(\dfrac{20}{2\times7}+\dfrac{20}{7\times12}+\dfrac{20}{12\times17}+\dfrac{20}{17\times22}+\dfrac{20}{22\times27}+\dfrac{20}{27\times32}\)
\(=4\times\left(\dfrac{5}{2\times7}+\dfrac{5}{7\times12}+\dfrac{5}{12\times17}+\dfrac{5}{17\times22}+\dfrac{5}{22\times27}+\dfrac{5}{27\times32}\right)\)
\(=4\times\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{32}\right)\)
\(=4\times\left(\dfrac{1}{2}-\dfrac{1}{32}\right)=4\times\dfrac{15}{32}=\dfrac{30}{16}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 6 2023
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
D
1
5 tháng 10 2017
\(A=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{502.507}\)
\(=2\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+...+\frac{5}{502.507}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(=tự tính\)
HU
3
XO
23 tháng 3 2021
\(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{97.102}=\frac{3}{5}\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{9}{97.102}\right)\)
\(=\frac{3}{5}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{97}-\frac{1}{102}\right)=\frac{3}{5}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{3}{5}.\frac{25}{51}=\frac{5}{17}\)
B
4
31 tháng 8 2023
\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{102-97}{97.102}\)
\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{97}-\dfrac{1}{102}\)
\(=\dfrac{1}{2}-\dfrac{1}{102}=\dfrac{25}{51}\)
31 tháng 8 2023
\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{97-92}{92.97}\)
\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{92}-\dfrac{1}{97}\)
\(=\dfrac{1}{2}-\dfrac{1}{97}=\dfrac{95}{194}\)
S
2 tháng 9 2017
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
10 tháng 4 2016
TỬ SỐ CỦA PHÂN SỐ THỨ 50 LÀ : (50-1) *1+3=52
MẪU SỐ CỦA PHÂN SỐ THỨ 50 LÀ: 52+1=53
PHÂN SỐ THỨ 50 CỦA DÃY LÀ : 52/53
ĐÁP SỐ: 52/53
không biết nói làm gì :3
Phân số thứ 50 của dãy số sẽ có dạng là:
\(\frac{3}{\left(2+5\times49\right)\left(2+5\times50\right)}=\frac{3}{247\times252}\)