\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

đơn giản quá!

Bạn có bt làm bài 5 ko?

Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)

a, \(-1\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{2}+2\dfrac{1}{6}\\ =-\dfrac{5}{3}+\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{13}{6}\\ =\dfrac{-5.4+3.3-1.6+13.2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

b, \(\dfrac{11}{50}\left(-17\dfrac{1}{2}\right)-\dfrac{11}{50}.82\dfrac{1}{2}\\ =\dfrac{11}{50}.\left(-17\dfrac{1}{2}-82\dfrac{1}{2}\right)=\dfrac{11}{50}.\left(-100\right)=-22\)

a) \(-1\dfrac{2}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(2\dfrac{1}{6}\)

=\(-\dfrac{5}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(\dfrac{13}{6}\)\()\)

=\(-\) \(\dfrac{20}{12}\) + \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) + \(\dfrac{26}{12}\)

= \((\)\(\dfrac{-20}{12}\) + \(\dfrac{26}{12}\) \()\) + \((\) \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) \()\)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)

= \(\dfrac{3}{4}\)

b)\(\dfrac{11}{50}\) \((\) \(-17\dfrac{1}{2}\) \()\) \(-\) \(\dfrac{11}{50}\) .\(82\dfrac{1}{2}\)

= \(\dfrac{11}{50}\) . \(-\dfrac{35}{2}\) \(-\) \(\dfrac{11}{50}\) . \(\dfrac{165}{2}\)

= \(\dfrac{11}{50}\). \((\) \(-\dfrac{35}{2}\) \(-\) \(\dfrac{165}{2}\) \()\)

=\(\dfrac{11}{50}\). \(-\)\(100\)

= \(-22\)

Chúc bạn học thật tốt nha !

Bài 1: Tìm x biết:

a) \(\dfrac{6}{5}-2\left|1-3x\right|=1\dfrac{2}{3}\)

\(2\left|1-3x\right|=\dfrac{6}{5}-1\dfrac{2}{3}\)

\(2\left|1-3x\right|=\dfrac{-7}{15}\)

\(\left|1-3x\right|=\dfrac{-7}{15}:2\)

\(\left|1-3x\right|=\dfrac{-7}{30}\)

\(\left|1-3x\right|\in N\) nhưng \(\dfrac{-7}{30}\notin N\)

\(\Rightarrow x=\varnothing\)

b) \(\left(2,8x+50\right):\dfrac{-3}{2}=51\)

\(\left(2,8x+50\right)=51.\dfrac{-3}{2}\)

\(2,8x+50=\dfrac{-153}{2}\)

\(2,8x=\dfrac{-153}{2}-50\)

\(2,8x=\dfrac{-253}{2}\)

\(x=\dfrac{-253}{2}:2,8\)

\(x=\dfrac{-1265}{28}\)

c) \(\dfrac{x-2}{-2}=\dfrac{x+4}{3}\)

\(\Rightarrow\left(x-2\right).3=-2.\left(x+4\right)\)

\(x.3-2.3=\left(-2\right).x+\left(-2\right).4\)

\(3x-6=\left(-2\right)x+\left(-8\right)\)

\(3x-\left(-2\right)x=6+\left(-8\right)\)

\(5x=-2\)

\(x=\left(-2\right):5\)

\(x=\dfrac{-2}{5}\)

d) \(4\left(3-2x\right)-5\left(x-1\right)=12\)

\(4.3-4.2x-5x+5.1=12\)

\(12-8x-5x+5=12\)

\(12+\left(-8\right)x+\left(-5\right)x+5=12\)

\(12+\left(-13\right)x+5=12\)

\(\left(-13\right)x=12-12-5\)

\(\left(-13\right)x=-5\)

\(x=\left(-5\right):\left(-13\right)\)

\(x=\dfrac{5}{13}\)

Bài 2: Chứng minh:

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\) (đpcm)

bài này có trong sách Nâng cao và Phát triển bạn nhé

a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)

=>2/5x=8/5

=>x=4

b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)

=>1/3x=-6

=>x=-18

c: =>2|x-1/3|=0,24-4/5=-0,56<0

Giúp với

Ta có:

P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\left(1+1+...+1\right)\)(có 49 chữ số 1)

P= \(\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)

P= \(\dfrac{50}{49}+\dfrac{50}{48}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

P= \(50.\left(\dfrac{1}{50}+\dfrac{1}{49}+...+\dfrac{1}{2}\right)\)

⇒ \(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}}{50.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)}\)

⇒ \(\dfrac{S}{P}=\dfrac{1}{50}\)

Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)

a: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5x3/2=9/10

b: \(\Leftrightarrow x\cdot2.8-50=34\)

=>2,8x=84

=>x=30

c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)

hay x=5/2

d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)

=>2x-3/4=41/4 hoặc 2x-3/4=-41/4

=>2x=44/4=11 hoặc 2x=-19/2

=>x=11/2 hoặc x=-19/4