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\(a,x-\frac{5}{6}:1\frac{1}{6}=0,125\)
\(x-\frac{5}{6}:\frac{7}{6}=\frac{1}{8}\)
\(x-\frac{5}{7}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{7}\) \(x=\frac{47}{56}\)
\(b,\left(1-\frac{2}{10}+x+\frac{1}{5}\right):\left(1\frac{1}{3}-\frac{2}{3}+3\frac{1}{3}\right)-1=1\frac{1}{2}\)
\(\left(1-\frac{1}{5}+x+\frac{1}{5}\right):\left(\frac{4}{3}-\frac{2}{3}+\frac{10}{3}\right)-1=\frac{3}{2}\)
\(\left(\frac{4}{5}+x+\frac{1}{5}\right):4=\frac{3}{2}+1\)
\(\left(1+x\right):4=\frac{5}{2}\)
\(1+x=\frac{5}{2}.4\)
\(1+x=10\)
\(x=10-1\)
\(x=9\)
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
S = \(\frac{1}{2}\)+ \(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{24}\)+\(\frac{1}{48}\)
S = \(\frac{24}{48}\)+ \(\frac{8}{48}\)+\(\frac{4}{48}\)+\(\frac{2}{48}\)+\(\frac{1}{48}\)
S = \(\frac{39}{48}\)= \(\frac{13}{16}\)
\(\Rightarrow\)S = \(\frac{13}{16}\)
Cho A = 1/2 + 1/6 + 1/12 + 1/20 + ... + 1/n
Biết A = 39/40, tìm n
Các bạn ghi cả lời giải ra nữa nhé !
Giải
Ta có: 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + ..... 1/n = 39/40
Đặt n = k x (k+1)
=> 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(k x (k+1)) = 39/40
=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/k - 1/(k+1) = 39/40
=> 1 - 1/(k+1) = 39/40
=> 1/(k+1) = 1 - 39/40
=> 1/(k+1) = 1/40
=> k+1 = 40
=> k = 39
Vậy n = k x (k+1) = 39 x (39+1) = 39x40 = 1560
ĐS: n= 1560
siêu đúng
Ta thấy:
1/2 = 1/1.2 = 1-1/2
1/6=1/2.3 = 1/2-1/3
1/12=1/3.4 = 1/3-1/4
1/20=1/4.5 = 1/4-1/5
1/30=1/5.6 = 1/5-1/6 ………..
1/n =1/(a-1).a = 1/(a-1) – 1/a (n = (a-1).a) (1)
Vậy
1/2+1/6+1/12+1/20+1/30+ ….. + 1/n = 1 – 1/a
Hay: 1 – 1/a = 39/40
1/a = 1 – 39/40 = 1/40
a = 40
Thay a vào (1) ta được:
n = (40-1) x 40 = 1560