Câu hỏi của Phương Uyên - Toán lớp 7 | Học trực tuyến

Ta có:

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)

Cảm ơn K tồn tại nhiều lắm nhưng ở bài của bn k có tick đúng vs bình luận.

2, a-b=ab => a=ab+b => a=b(a+1)

thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1

Theo bài ra ta có: a:b=a-b

=> a+1=a-b

=>-b=1

=> b=-1

Thay b=-1 vào a-b=ab ta có : a-(-1)=-a

=> a +1=-a

=>a=-1/2

Vậy a=-1/2. b=-1

Ta có :

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+......+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+.....+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+......+\dfrac{1}{50}\)

Vậy ...

Đặt:

\(PHUCDZ=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(PHUCDZ=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(PHUCDZ=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\)

\(PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)

\(PHUCDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

Đặt \(PHUCMAXDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

\(PHUCDZ=PHUCMAXDZ\) vậy ta có \(đpcm\)

Ta có : \(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\) + \(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{49.50}\)

= (1-\(\dfrac{1}{2}\)) + (\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) + (\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)) + ... + (\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\))

= (1+\(\dfrac{1}{3}\) +\(\dfrac{1}{5}\)+....+\(\dfrac{1}{49}\)) - ( \(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{50}\))

=(1+\(\dfrac{1}{3}\)+\(\dfrac{1}{5}\)+...+\(\dfrac{1}{49}\)) - 2(\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{50}\))

= (1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{50}\)) - (1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{25}\))

=\(\dfrac{1}{26}\)+\(\dfrac{1}{27}\)+...+\(\dfrac{1}{50}\) (đpcm)

a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

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

\(\Rightarrowđpcm\)

d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)

\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)

\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)

\(\Rightarrowđpcm\)

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

\(\Rightarrowđpcm\)

Đặt A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.......+\dfrac{1}{3^{99}}\)

=> 3A=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+..........+\dfrac{1}{3^{98}}\)

=> 3A-A= 1-\(\dfrac{1}{3^{99}}\)

=> A=\(\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)

=> A<1/2

Vậy A<1/2

Ta có:

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)

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

\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)

\(B\text{=}\dfrac{247}{300}\)

Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)

Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)

\(\Rightarrowđpcm\)

Dạng hay :v

Ta có:
\(A = \dfrac{1}{1.2} + \dfrac{1}{3.4} +...+ \dfrac{1}{49.50}\)
\(=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{49})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50})\)
\(=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50})-2.(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50})\)
\(=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50})-(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25})\)
\(=>A=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50} (1)\)
Ta lại có:
\(B = \dfrac{1}{26.50} + \dfrac{1}{27.49} +...+ \dfrac{1}{50.26}\)
\(=>38B=\dfrac{38}{26.50}+\dfrac{38}{27.49}+...+\dfrac{38}{50.26}\)
\(=>38B=\dfrac{76}{26.50}+\dfrac{76}{27.49}+...+\dfrac{38}{38.38}\)
\(=>38B=\dfrac{1}{26}+\dfrac{1}{50}+\dfrac{1}{27}+\dfrac{1}{49}+...+\dfrac{1}{38}\)
\(=>38B=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50} (2)\)
Từ (1)(2):
\(=>A = 38B\)
\(=>A-38B=0\)

Câu hỏi của Phương Uyên - Toán lớp 7 | Học trực tuyến

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