CMR:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
chứng minh rằng
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
Câu hỏi của Phương Uyên - Toán lớp 7 | Học trực tuyến
CMR : \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
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.
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{49}+\dfrac{1}{50}\)
1.Chứng minh rằng:
a) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
b) Cho A = \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
Chứng minh \(\dfrac{7}{12}< A< \dfrac{5}{6}\)
2. Tìm a, b \(\in\) Q, biết
a - b = a.b = a : b
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
8 Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) ; b) \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100}{100!}\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
d) \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}< \dfrac{1}{2}\)
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
Nhờ các bạn giúp đỡ bài này ạ:
Tính tổng:
A = \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+...+\dfrac{1}{49.50}\)
Cảm ơn các bạn.
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{49}-\dfrac{1}{50}\)
=\(\dfrac{1}{1}-\dfrac{1}{50}\)=\(\dfrac{49}{50}\)
Bài 1: Cho M= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)+...+\(\dfrac{1}{50}\)
Chứng minh M = \(\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
MÌNH CẢM ƠN ĐÃ GIÚP CHO MÌNH~
Câu hỏi của Phương Uyên - Toán lớp 7 | Học trực tuyến
mình ko có thời gian
bạn tự xem nhé
3. tính:
A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
A = 1/1.2 +1/2.3 +1/3.4 +...+1/49.50
A = 1 +1/2 -1/2+1/3-1/3+1/4-...-1/49 +1/50
A = 1 - 1/50
A=49/50
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
CMR
\(\dfrac{7}{12}< A< \dfrac{5}{6}\)
A= \(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
Ta có :
\(\dfrac{7}{12}=\dfrac{7\cdot50}{12\cdot50}=\dfrac{350}{600}\)
\(\dfrac{99}{100}=\dfrac{99\cdot6}{100\cdot6}=\dfrac{594}{600}\)
\(\dfrac{5}{6}=\dfrac{5\cdot100}{6\cdot100}=\dfrac{500}{600}\)
Chỗ này hình như bn viết sai đề nha
QĐMS lên phải là \(\dfrac{7}{12}< \dfrac{5}{6}< A\) chứ
Bn xem lại đề ik nha ☺
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