\(\dfrac{1}{49}+\dfrac{49}{1}\)

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26 tháng 4 2017

\(\dfrac{1}{49}+\dfrac{49}{1}\) \(=\) \(\dfrac{1}{49}+49=49\dfrac{1}{49}\)

16 tháng 3 2018

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

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

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

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

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

24 tháng 3 2017

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

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

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

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

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

21 tháng 3 2017

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

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

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

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

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

4 tháng 5 2018

Giúp vớikhocroi

18 tháng 7 2018

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 \(\dfrac{S}{P}=\dfrac{1}{50}\)

11 tháng 5 2017

\(Q=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\\ =\dfrac{1}{49}+1+\dfrac{2}{48}+1+\dfrac{3}{47}+1+...+\dfrac{47}{3}+1+\dfrac{48}{2}+1+1\\ =\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{3}+\dfrac{50}{2}+\dfrac{50}{50}\\ =50\cdot\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+...+\dfrac{1}{3}+\dfrac{1}{2}\right)\\ =50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

\(\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}}{50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

12 tháng 5 2017

bn thiếu dấu ngoặc ở phép thứ 2 rồi

8 tháng 4 2017

à mk nhầm thay 50 * b thành a nha

23 tháng 5 2017

Đề sai à???

Đáng ra phải là \(\dfrac{A}{B}\) chứ???

Với cả nếu muốn CM biểu thức ko là số tự nhiên thì chỉ cần có 1 biểu thức thui chứ nhỉ, cần j 2???

5 tháng 5 2018

P = 1/49+2/48+3/47+...+48/2+49/1

Cộng 1 váo mỗi p/s trong 48 p/s đầu , trừ p/s cuối đi 48 ta được

P=(1/49+1)+(2/48+1)+...+(48/2+1)+1

P= 50/49+50/48+....+50/2+50/50

Đưa ps cuối lên đầu

P=50/50+50/49+50/48+...+50/2

=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)

=50S

=> S/P=1/50

mk nhầm các bn thay50 * b thành A nha

8 tháng 8 2018

1) \(x:\dfrac{2}{3}=150\)

\(\Leftrightarrow x=150.\dfrac{2}{3}\)

\(\Leftrightarrow x=100\).

2) \(\dfrac{35}{9}:x=\dfrac{35}{6}\)

\(\Leftrightarrow x=\dfrac{35}{9}:\dfrac{35}{6}\)

\(\Leftrightarrow x=\dfrac{2}{3}\).

3) \(\dfrac{49}{7}:x=\dfrac{49}{5}\)

\(\Leftrightarrow x=\dfrac{49}{7}:\dfrac{49}{5}\)

\(\Leftrightarrow x=\dfrac{5}{7}\).

4) \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{5}=0\)

\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{78}{5}=0\)

\(\Leftrightarrow\left\{\dfrac{-35}{18}+x\right\}:\dfrac{78}{5}=1-0\)

\(\Rightarrow\dfrac{-35}{18}+x=1.\dfrac{78}{5}\)

\(\Leftrightarrow\dfrac{-35}{18}+x=\dfrac{78}{5}\)

\(\Rightarrow x=\dfrac{1579}{90}\).

8 tháng 8 2018

Gọi a,b,c.. cho dễ nhé.Thớt vui tính quá, dấu phẩy cũng không viết hộ con dân =)))

a, \(x:\dfrac{2}{3}=150\)

\(\Leftrightarrow x=150.\dfrac{2}{3}\)

\(\Leftrightarrow x=100\)

Vậy...

b, \(\dfrac{35}{9}:x=\dfrac{35}{6}\)

\(\Leftrightarrow x=\dfrac{35}{9}:\dfrac{35}{6}\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy...

c, \(\dfrac{49}{7}:x=\dfrac{49}{5}\)

\(\Leftrightarrow x=\dfrac{49}{7}:\dfrac{49}{5}\)

\(\Leftrightarrow x=\dfrac{5}{7}\) Vậy...

d, \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{4}=0\)

\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{63}{4}=0\)

\(\Leftrightarrow1-\left\{x-\dfrac{35}{18}\right\}:\dfrac{63}{4}=0\)

\(\Leftrightarrow1-\left(\dfrac{\left(18x-35\right).4}{18.63}\right)=0\)

\(\Leftrightarrow1-\left(\dfrac{72x-140}{1134}\right)=0\)

\(\Leftrightarrow1-\dfrac{72x-140}{1134}=0\)

\(\Leftrightarrow\dfrac{1134-72x+140}{1134}=0\)

\(\Leftrightarrow1274-72x=0\)

\(\Leftrightarrow72x=1274\)

\(\Leftrightarrow x=\dfrac{637}{36}\)

Vậy...