1.\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{23}-\frac{4}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

2. Đặt \(A=\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+\frac{3}{374}+\frac{3}{594}+\frac{3}{864}\)

\(\Rightarrow A=\frac{3}{2.7}+\frac{3}{7.12}+...+\frac{3}{27.32}\)

\(\Rightarrow5A=3.\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{27.32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{27}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\frac{15}{32}=\frac{45}{32}\Rightarrow A=\frac{45}{32}:5=\frac{9}{32}\)

3. Đặt \(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\)

\(\Rightarrow3S=\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow S=\frac{9}{20}:3=\frac{3}{20}\)

Câu 1:

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

Câu 1:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

Áp dụng tính chất \(\frac{b}{a\left[a+b\right]}=\frac{1}{a}-\frac{1}{a+b}\), ta có:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

Câu 2 tương tự nhưng phải phân

\(\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+...+\frac{3}{864}=\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}\)

Cái này áp dụng công thức \(\frac{a}{b\left[b+c\right]}=\frac{a}{c}\left[\frac{1}{b}-\frac{1}{b+c}\right]\), ta có:

\(\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{27}-\frac{1}{32}\right]\)

\(=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{32}\right]=\frac{3}{5}\cdot\frac{15}{32}=\frac{9}{32}\)

Câu 3:

tương tự quy laautj mẫu là 2.5; 5.8 ....

Câu 4: qL mẫu là 1.7; 7.13; ....

Câu 5: \(=\left[1-\frac{1}{2}\right]+\left[1-\frac{1}{6}\right]+...+\left[1-\frac{1}{110}\right]\)

\(=\left[1-\frac{1}{1\cdot2}\right]+\left[1-\frac{1}{2.3}\right]+\left[1-\frac{1}{3\cdot4}\right]+...+\left[1-\frac{1}{10.11}\right]\)

\(=10-\frac{9}{10}=\frac{91}{10}\)

a, 72/65

lời giải phần a, đâu ? mình cần cả lời giải và đáp số.

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

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

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

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

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

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

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

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

\(#Wendy.Dang\)

x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

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