Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
\(a\)) Mình giải theo cách khác:
Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)
Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
Bài 2:
\(A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
\(B=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)
\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)
2 phần dưới không liên quan gì đến tính chất trên
a) \(A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{20-17}{17.20}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
b) \(B=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{106-101}{101.106}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{106}\right)\)
\(B=5.\left(1-\frac{1}{106}\right)=\frac{525}{106}\)
`A = ( 5^2 )/( 1*6)+(5^2)/(6*11)+.....+(5^2)/(26*31)`
`= 5*( 5/( 1*6)+ 5/(6*11)+.....+5/(26*31))`
`= 5*( 1 - 1/6 + 1/6 - 1/11 +....+1/26 - 1/31 )`
`= 5*( 1 - 1/31 )`
`= 5 * 30/31 = 150/31`
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
\(A=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)
\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)
\(B=\frac{5^2}{1.6}+\frac{5^2}{6.11}+.....+\frac{5^2}{26.31}\)
\(B=\frac{5.5}{1.6}+\frac{5.5}{6.11}+.....+\frac{5.5}{26.31}\)
\(B=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+.......+\frac{5}{26.31}\right)\)
\(B=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+......+\frac{1}{26}-\frac{1}{31}\right)\)
\(B=5.\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(B=\frac{5.30}{31}\)
\(B=\frac{150}{31}\)
A=\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+....+\frac{5^2}{26.31}\)
=>A=5.(\(\frac{5}{1.6}+\frac{5}{6.11}+....+\frac{5}{26.31}\))
=>A=5.(\(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\))
=>A=5.(\(\frac{1}{1}-\frac{1}{31}\))
=>A=5.\(\frac{30}{31}\)
=>A=\(\frac{150}{31}\)
=>A>1( vì tử của A lớn hơn mẫu )
a, gọi ƯCLN(14n+3;21n+5)=d
=> \(\left\{{}\begin{matrix}14n+3\\21n+5\end{matrix}\right.\)⋮d =>\(\left\{{}\begin{matrix}3\left(14n+3\right)\\2\left(21n+5\right)\end{matrix}\right.\)⋮d=>\(\left\{{}\begin{matrix}42n+9\\42n+10\end{matrix}\right.\)⋮d
=>(42n+10)-(42n+9)⋮d
=>1⋮d
=>d=1
Do ƯCLN của 14n+3 ; 21n+5 là 1
=> 2 số trên là hai số nguyên tố cùng nhau
=>hai số đó nếu chia cho nhau thì sẽ ko chia hết
=> hai số đó khi biểu diễn ở dạng phân số thì sẽ thành phân số tối giản
a, A = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\)
A = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...\dfrac{1}{17}-\dfrac{1}{20}\)
A = \(\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{10}{20}-\dfrac{1}{20}=\dfrac{9}{20}\)
Lần sau nếu có bài dạng như thế này bạn hãy làm theo quy tắc sau nha: \(\dfrac{m}{b.\left(b+m\right)}=\dfrac{m}{b}-\dfrac{m}{b+m}\)
Tick cho mk vs. thank!!!
b, B = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
B = \(\dfrac{5^2}{5}.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
B = \(5.\left(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
B = \(5.\left(\dfrac{1}{1}-\dfrac{1}{31}\right)\)
B = \(5.\dfrac{30}{31}=\dfrac{150}{31}\)
Tick cho mk nha! please đó!
