Chứng minh rằng
A= \(\frac{1}{2}\)+ \(\frac{1}{33}\)+\(\frac{1}{34}\)+\(\frac{1}{35}\)+\(\frac{1}{51}\)+\(\frac{1}{53}\)+\(\frac{1}{55}\)+\(\frac{1}{57}\)+\(\frac{1}{59}\)< \(\frac{7}{10}\)
giúp mk nha
mk cần gấp
a) \(\frac{59-x}{41}+\frac{57-x}{43}=\frac{55-x}{45}+\frac{53+x}{47}+\frac{51-x}{49}=-5\)
b) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
Giúp mình với!
a, <=> (59-x/41 + 1) + (57-x/43 + 1) + (55-x/45 + 1) + (53-x/47 + 1) + (51-x/49 + 1) = 0
<=> 100-x/41 + 100-x/43 + 100-x/45 + 100-x/47 + 100-x/49 = 0
<=> (100-x).(1/41+1/43+1/45+1/47+1/49) = 0
<=> 100-x=0 ( vì 1/41+1/43+1/45+1/47+1/49 > 0 )
<=> x=100
Vậy x = 100
b, <=> 2-x/2016 + 1 = (1-x/2017 + 1) + (1 - x/2018)
<=> 2018-x/2016 = 2018-x/2017 + 2018-x/2018
<=> 2018-x/2016 - 2018-x/2017 - 2018-x/2018 = 0
<=> (2018-x).(1/2016-1/2017-1/2018) = 0
<=> 2018-x=0 ( vì 1/2016-1/2017-1/2018 khác 0 )
<=> x=2018
Vậy x=2018
Tk mk nha
Chứng minh : \(\frac{1}{1}-\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{5}-\frac{1}{8}+\frac{1}{13}-\frac{1}{21}+\frac{1}{34}-\frac{1}{55}+...< \frac{3}{10}\).
\(A=\frac{\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+...+\frac{1}{149}}{\frac{1}{101.109}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)
Chứng minh \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}< 1\)
Ta có:\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+............+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+........+\frac{1}{100}\right)\)
\(>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{1}{2}\)
\(\left(\frac{1}{51}+\frac{1}{52}+..........+\frac{1}{75}\right)+\left(\frac{1}{76}+........+\frac{1}{100}\right)\)
\(< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}< 1\)
\(\Rightarrowđpcm\)
Chứng minh :
\(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+....+\frac{1}{100}< 1\)
Vì mọi phân số của tổng đều nhỏ hơn 1 nên tổng đó nhỏ hơn 1.
k nha
Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
\(\frac{\frac{3}{51}-\frac{12}{57}+\frac{33}{59}}{\frac{5}{51}-\frac{20}{57}+\frac{55}{59}}\)
\(=\frac{3.\left(\frac{1}{51}-\frac{4}{57}+\frac{11}{59}\right)}{5.\left(\frac{1}{51}-\frac{4}{57}+\frac{11}{59}\right)}=\frac{3.1}{5.1}=\frac{3}{5}\)
Chứng minh rằng:
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)
Trả lời
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
\(\Leftrightarrow\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)
\(\Leftrightarrow\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}\)
\(\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\left(đpcm\right)\)
1/2 lớn hơn
vì phân số 1/2 có mẫu số nhỏ hơn các phân số kia nên phân số 1/2 sẽ lớn hơn các phân số kia
Chứng minh rằng:
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\le\frac{1}{2}\)