\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)

\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)

\(a\)) 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:

\(A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)

\(B=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)

\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)

Bài này mình làm rồi bạn NGUYỆT ạ

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) \(B=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{106-101}{101.106}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{106}\right)\)
\(B=5.\left(1-\frac{1}{106}\right)=\frac{525}{106}\)

Có liên quan đó bạn!

\(A=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)

\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)

bằng \(\dfrac{5}{3}\)

\(B=\frac{5^2}{1.6}+\frac{5^2}{6.11}+.....+\frac{5^2}{26.31}\)

\(B=\frac{5.5}{1.6}+\frac{5.5}{6.11}+.....+\frac{5.5}{26.31}\)

\(B=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+.......+\frac{5}{26.31}\right)\)

\(B=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+......+\frac{1}{26}-\frac{1}{31}\right)\)

\(B=5.\left(\frac{1}{1}-\frac{1}{31}\right)\)

\(B=\frac{5.30}{31}\)

\(B=\frac{150}{31}\)

x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

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