1, 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
2, 1/6 + 5/6.7 + 1/7.8 + 5/8.9 + ... + 1/24.25
3, 22/5.7 + 22/7.9 + 22/9.11 + ... + 22/39.41
Tính nhanh
1, A = 1/1.2 + 1/2.3 + 1/3.4 + 1/3.4 + ... + 1/49.50
2, B = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39
3, C = 5^2/1.6 + 5^2/6.11 + ... + 5/26.31
4, D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
x + 25 = 64
x = 64 - 25
x = 39
Vậy x = 39
B= 1 trên 3.4 + 1 trên 4.5 + 1 trên 5. 6 + 1 trên 6.7 +1 trên 7.8
C= 1 trên 4.5 +1 trên 5.6 + 1 trên 6.7 + 1 trên 7.8 +1 trên 8.9
D= 2 3.5 + 2 5.7 + 2 7.9 + 2 9.11 +2 11.13
E = 3 1.4 + 3 4.7 + 3 7.10 + 3 10.13
Ta có: \(B=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}\)
\(=\frac{5}{24}\)
Vậy \(B=\frac{5}{24}\)
Ta có: \(C=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{4}-\frac{1}{9}\)
\(=\frac{5}{36}\)
Vậy \(C=\frac{5}{36}\)
Ta có: \(C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}\)
\(=\frac{10}{39}\)
Vậy \(D=\frac{10}{39}\)
1/7+1/91+1/247+1/475+1/775+1/1147
\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}.\dfrac{36}{37}\)
\(=\dfrac{6}{37}\)
\(#Wendy.Dang\)
giá trị của tổng : 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147 = ?
= 6/37 nha
đúng 10000000000000000000000000% đó nếu có sai thì trách cái máy tính của mk nha vì k pải mk tính mờ là nó tính....
tk mk nha đg âm 439 kìa....
T_T
tính D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
=\(\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
=\(\frac{1}{6}.\frac{6}{1.7}+\frac{1}{6}.\frac{6}{7.13}+\frac{1}{6}.\frac{6}{13.19}+\frac{1}{6}.\frac{6}{19.25}+\frac{1}{6}.\frac{6}{25.31}+\frac{1}{6}.\frac{6}{31.37}\)
=\(\frac{1}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)
=\(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
=\(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{37}\right)\)
=\(\frac{1}{6}\left(\frac{37}{37}-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)
\(D=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)
\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
=1/1*7+1/7*13+1/13*19+1/19*25+1/25*31+1/31*37
=1/6(6/1*7+6/7*13+...+6/31*37)
=1/6(1-1/7+1/7-1/13+...+1/31-1/37)
=1/6*36/37=6/37
Tính
1\7+1\91+1\247+1\475+1\775+1\1147
ta làm theo cách sau đây :
▬ Min của x² + y²:
Áp dụng bđt bunhiacôpxki cho cặp số x²,y² và 1,1 ta có:
...........(x² + y²)(1 + 1) ≥ (x + y)² ≥ 2² = 4
....<=> (x² + y²) ≥ 2
=> Min x² + y² = 2 <=> x = y = 1
▬ Min của x³ + y³:
Áp dụng bđt Cauchy cho 2 số dương a² và b² ta có:
............x² + y² ≥ 2.x.y
.....<=> -2.x.y ≥ x² + y² ≥ 2
.....<=> -.x.y ≥ 1
Ta có: x³ + y³ = (x + y).(x² + y² - x.y)
=> x³ + y³ ≥ 2.(2 + 1) ≥ 6
=> MIn x³ + y³ = 6 <=> x = y = 1
▬ Min của x^4 + y^4
Áp dụng bđt bunhiacôpxki cho cặp số x^4,y^4 và 1,1 ta có:
...........(x^4 + y^4)(1 + 1) ≥ (x² + y)² ≥ 2² = 4
......=> (x^4 + y^4) ≥ 2
=> Min x^4 + y^4 = 2 <=> x = y = 1
hoặc bạn có thể :
A=1/7 +1/91 +1/247 + 1/475 + 1/775 + 1/1147
A=1/(1.7)+1/(7.13)+1/(13.19)+...+1/(31...
A=(1/6)*( 1 - 1/7 + 1/7 - 1/13 +... +1/31-1/37)
A=(1/6)*(1-1/37)
A=(1/6)*(36/37)
A=6/37
.
B= 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/45
B= 2/(2.3) + 2/(3.4) + 2/(4.5) + ... + 2/(9.10)
B= 2(1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
B= 2(1/2-1/10)
B= 4/5
= 1/6 x (6/1x7+6/7x13+6/13x19+6/19x25+6/25x31+6/31x37)
= 1/6 x (1-1/7+1/7-1/13+1/13-1/19+1/19-1/25+1/25-1/31+1/31-1/37)
= 1/6 x (1-1/37)
= 1/6 x 36/37
= 6/37
Tính nhanh
a, M = 3/2 + 3/6 + 3/13 + 3/20 + 3/30 + 3/42
b, N = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
lời giải phần a, đâu ? mình cần cả lời giải và đáp số.
-1/91+-1/247+-1/475+-1/775+-1/1147