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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
D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
=\(\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
=\(\frac{1}{6}.\frac{6}{1.7}+\frac{1}{6}.\frac{6}{7.13}+\frac{1}{6}.\frac{6}{13.19}+\frac{1}{6}.\frac{6}{19.25}+\frac{1}{6}.\frac{6}{25.31}+\frac{1}{6}.\frac{6}{31.37}\)
=\(\frac{1}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)
=\(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
=\(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{37}\right)\)
=\(\frac{1}{6}\left(\frac{37}{37}-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)
\(D=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)
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
Ừk
7.
\(G=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\\ =\dfrac{1}{3}-\dfrac{1}{13}\\ =\dfrac{13}{39}-\dfrac{3}{39}\\ =\dfrac{10}{39}\)
8.
\(H=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{755}+\dfrac{1}{1147}\\ =\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\dfrac{1}{19\cdot25}+\dfrac{1}{25\cdot31}+\dfrac{1}{31\cdot37}\\ =\dfrac{1}{6}\cdot\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+\dfrac{6}{19\cdot25}+\dfrac{6}{25\cdot31}+\dfrac{6}{31\cdot37}\right)\\ =\dfrac{1}{6}\cdot\left(\dfrac{1}{1}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\\ =\dfrac{1}{6}\cdot\left(1-\dfrac{1}{37}\right)\\ =\dfrac{1}{6}\cdot\dfrac{36}{37}\\ =\dfrac{6}{37}\)
a: =(2/7-2/7)(-4/7-5/9)=0
b:
Sửa đề: 9/13*(-12/17)+9/13*29/27
=9/13(-12/17+29/17)
=9/13*17/17=9/13
c: \(=\dfrac{1}{7}\left(4+\dfrac{6}{7}+\dfrac{8}{7}\right)=\dfrac{1}{7}\cdot6=\dfrac{6}{7}\)
d: =7/10(5/7+9/7+8/7+13/7)
=5*7/10=7/2
\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+...+\frac{1}{31\cdot37}\)
\(=\frac{1}{6}\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\frac{6}{13\cdot19}+...+\frac{6}{31\cdot37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...-\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\frac{36}{37}=\frac{6}{37}\)
Tổng cần tính bằng:\(\frac{1}{1.7}\)+\(\frac{1}{7.13}\)+\(\frac{1}{13.19}\)+\(\frac{1}{19.25}\)+\(\frac{1}{25.31}\)+\(\frac{1}{31.37}\)=(\(\frac{1}{1}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{13}\)+...+\(\frac{1}{31}\)\(\frac{1}{37}\)):3 =(\(1\)-\(\frac{1}{37}\)):3=\(\frac{12}{37}\)
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
=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