a) Đặt \(d=\left(2n+7,5n+17\right)\)

Suy ra 

\(\hept{\begin{cases}2n+7⋮d\\5n+17⋮d\end{cases}}\Rightarrow5\left(2n+7\right)-2\left(5n+17\right)=1⋮d\Rightarrow d=1\)

Suy ra đpcm. 

b) \(A=\frac{2}{1.5}+\frac{3}{5.11}+\frac{4}{11.19}+\frac{5}{19.29}+\frac{6}{29.41}\)

\(=\frac{1}{2}\left(\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\right)\)

\(=\frac{1}{2}\left(\frac{5-1}{1.5}+\frac{11-5}{5.11}+\frac{19-11}{11.19}+\frac{29-19}{19.29}+\frac{41-29}{29.41}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{41}\right)=\frac{20}{41}\)

\(B=\frac{1}{1.4}+\frac{2}{4.10}+\frac{3}{10.19}+\frac{4}{19.31}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}\right)\)

\(=\frac{1}{3}\left(\frac{4-1}{1.4}+\frac{10-4}{4.10}+\frac{19-10}{10.19}+\frac{31-19}{19.31}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{31}\right)=\frac{10}{31}\)

\(A\div B=\frac{20}{41}\div\frac{10}{31}=\frac{62}{41}\)

.............       m,m.lkjm,lkijnm,klkijm,

ai bt

mới lớp 5 à

thế còn trả lời

em lớp 5 mà còn bt á

2xy – 10x + y = 16

⇒ ( 2x + 1 ) – 10x = 16

⇒  2x + 1 = 0

⇒ 2x = -1

⇒ 2 ( x – 5 ) = 0

⇒ y = 5

Chú ý giải bằng lập bảng