0,1621963429

\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}\)

\(=\frac{1}{7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)

\(=\frac{1}{6}\left(1-\frac{1}{37}\right)\)

\(=\frac{1}{6}.\frac{36}{37}\)

\(=\frac{6}{37}\)

\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}\)

\(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)

\(=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{35.37}\right)\)

\(=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac{1}{6}.\left(1-\frac{1}{37}\right)\)

\(=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)

~ Hok tốt ~

\(\frac{1}{6}\left(\frac{6}{7}+\frac{6}{91}+\frac{6}{247}+\frac{6}{475}+\frac{6}{755}+\frac{6}{1147}\right)\)

\(=\frac{1}{6}\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\frac{6}{13\cdot19}+\frac{6}{19\cdot25}+\frac{6}{25\cdot31}+\frac{6}{31\cdot37}\right)\)

\(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac{1}{6}\left(1-\frac{1}{37}\right)\)

\(\frac{1}{6}\left(\frac{37}{37}-\frac{1}{37}\right)=\frac{1}{6}\cdot\frac{36}{37}=\frac{1\cdot6\cdot6}{6\cdot37}=\frac{6}{37}\)

\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

\(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}.\dfrac{36}{37}\)

\(=\dfrac{6}{37}\)

\(#Wendy.Dang\)

= 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

lấy máy tính mà tính

bằng 5/31 nha bạn

D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147

=\(\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)

=\(\frac{1}{6}.\frac{6}{1.7}+\frac{1}{6}.\frac{6}{7.13}+\frac{1}{6}.\frac{6}{13.19}+\frac{1}{6}.\frac{6}{19.25}+\frac{1}{6}.\frac{6}{25.31}+\frac{1}{6}.\frac{6}{31.37}\)

=\(\frac{1}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)

=\(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)

=\(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{37}\right)\)

=\(\frac{1}{6}\left(\frac{37}{37}-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)

\(D=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

\(=\frac{1}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)

\(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac{1}{6}\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)

kết quả là 6/37

=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

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