Chứng tỏ rằng: 1/1.2+1/2.3+...+1/49.50=1/26+1/27+...+1/50
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Chứng tỏ rằng: 1/1.2+1/2.3+...+1/49.50=1/26+1/27+...+1/50
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
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1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
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Chứng tỏ :
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 = 1/26 + 1/27 + .. + 1/ 50
1/1.2+1/2.3+...+1/49.50:1/26+1/27+...+1/50
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+...+\left(\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
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Chứng minh rằng:
a) 1.2 - 1 phần 2! + 2.3 -1 phần 3! + 3.4 -1/4! + ... + 99.100 -1 /100! < 2
b) 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + ... + 1/50
Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
A=1/1.2+1/3.4+.....+1/49.50
=1-1/2+1/3-1/4+...+1/49-1/50=(1+1/3+1/5+...+1/49) - (1/2+1/4+1/6+...+1/50)
=(1+1/3+1/5+...+1/49)+(1/2+1/4+1/6+...+1/50)-2.(1/2+1/4+1/6+...+1/50)
=(1+1/2+1/3+1/4+...+1/49+1/50) - (1+1/2+1/3+...1/25)
=1/26+1/27+...1/50
Vậy .........
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Chứng minh rằng: (1/26+1/27+...+1/50)÷(1/1.2+1/3.4+...+1/49.50)=1
Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
Khi đó : \(\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)=1\) (đpcm)
Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
Khi đó \(\frac{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}}=\frac{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}=1\left(\text{đpcm}\right)\)
CMR: 1/1.2+1/2.3+1/3.4+....+1/49.50=1/26+1/27+.....+1/49+1/50
1/1.2 + 1/2.3 + ...... + 1/49.50
= 1/1 - 1/2 + 1/2 - - .... - 1/50 = 1 - 1/50 = 49/50
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Chứng Minh Rằng: 1/1.2+1/3.4/1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{49}+\dfrac{1}{50}\)
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