Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mk ko viết đề nha:
a)=(1/2+1/2)+ (1/3+2/3)+(1/4 +3/4)+(1/5+4/5)
=(2/2 + 3/3) + (4/4 + 5/5)
=2/2 + 4/2
=6/4
À mik lm đc câu a thui nha!Nhưng k cho mik vs nhé!
Lời giải:
$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$
$\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$
$1-\frac{1}{4}+\frac{1}{x\times (x+1)}=\frac{1}{2}$
$\frac{1}{x\times (x+1)}=\frac{-1}{4}$ (đây là số âm lớp 4 chưa học). Bạn xem lại đề.
1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
= 1/1-1/6
=6/6-1/6
= 5/6
a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)
\(=1-\dfrac{1}{256}\)
\(=\dfrac{255}{256}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=1-\dfrac{1}{14}\)
\(=\dfrac{13}{14}\)
c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)
\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)
\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)
\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
\(=\dfrac{1}{15}-\dfrac{1}{90}\)
\(=\dfrac{6}{90}-\dfrac{1}{90}\)
\(=\dfrac{5}{90}=\dfrac{1}{18}\)
`1 / 2 + 1 / 3 - 1 / 3 - 1 / 4`
`= ( 1 / 2 - 1 / 4 ) + ( 1 / 3 - 1 / 3 )`
`= ( 2 / 4 - 1 / 4 ) + 0`
`= 1 / 4`