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\(S=7(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{63}) \)
\(S=7(\frac{1}{3}-\frac{1}{63})\)
\(S=7(\frac{21}{63}-\frac{1}{63}) \)
\(S=7.\frac{20}{63}\)
\(S=\frac{20}{9}\)
Do đó:\(S<\frac{5}{2}\)
S=\(\frac{2.7}{3.5}+\frac{2.7}{5.7}+\frac{2.7}{7.9}+....+\frac{2.7}{61.63}\)và\(\frac{5}{2}\)
S=7.(\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....-\frac{1}{63}\)) và\(\frac{5}{2}\)
S=7.(\(\frac{1}{3}-\frac{1}{63}\)) và\(\frac{5}{2}\)
S=7.\(\frac{20}{63}\)và\(\frac{5}{2}\)
=>S=\(\frac{20}{9}\)so với \(\frac{5}{2}\)
=>S=\(\frac{40}{18}\)và\(\frac{45}{18}\)
=>S<\(\frac{5}{2}\)
M=1/2{1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99}
=1/2{1/3-1/99}
=1/2*32/99
=16/99
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
Thay a,b,c lần lượt vào biểu thức...
Tính được kết quả:
a) A= \(-\frac{7}{10}\)
b) B= \(-\frac{2}{7}\)
c) C= 0
M=1+1/2^2+1/3^2+1/4^2+...+1/10^2>1+1/2*3+1/3*4+1/4^5+...+1/10*11
M>1+1/2-1/3+1/4-1/4+1/5-...-1/11
M>1+1/2-1/11
M>1+9/22
M>31/22
vì 31/22>4/3 nên M>4/3
a: \(B=\left(-\dfrac{1}{5}-\dfrac{5}{7}+\dfrac{-3}{35}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\dfrac{1}{41}\)
\(=\dfrac{-7-25-3}{35}+\dfrac{3+2+1}{6}+\dfrac{1}{41}=\dfrac{42}{41}-1=\dfrac{1}{41}\)
Không chép lại đề nhé
Ta có:
P=\(\frac{50-49}{49}+\frac{50-48}{48}+...+\frac{50-2}{2}+\frac{50-1}{1}\)
P=\(\frac{50}{49}-\frac{49}{49}+\frac{50}{48}-\frac{48}{48}+...+\frac{50}{2}-\frac{2}{2}+\frac{50}{1}-\frac{1}{1}\)
P=\(\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+\frac{50}{1}-\left(\frac{49}{49}+\frac{48}{48}+...+\frac{2}{2}+\frac{1}{1}\right)\)
P=\(50\cdot\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)+50-49\) (chỗ này gộp nha)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{48}+\frac{1}{49}\right)+1\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)+\frac{50}{50}\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)
=>P=50S
=>\(\frac{S}{P}=\frac{S}{50S}=\frac{1}{50}\)
Vừa nãy mình nói nhầm, Sorry.
a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm
b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)
= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{101.103}\)
\(=>A=\frac{3}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\right)\)
\(=>A=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(=>A=\frac{3}{2}.\left(1-\frac{1}{103}\right)=\frac{3}{2}.\frac{102}{103}=\frac{153}{103}>1\) (vì 153>103)
Vậy A>1
sorry,dòng thứ 2 sửa lại:\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{3}{101.103}\right)\) nhé!
Bạn xem lời giải của mình nhé:
Giải:
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{101.103}\\ =\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\\ =\frac{3}{2}.\left(1-\frac{1}{103}\right)\\ =\frac{3}{2}.\frac{103-1}{103}=\frac{3}{2}.\frac{102}{103}=\frac{153}{103}=1\frac{50}{103}\)
Chúc bạn học tốt!
2/3 A = 2/3. ( 3/1.3 + 3/3.5 + 3/5.7 +...+ 3/101.103 )
2/3 A= 2/1.3 + 2/3.5 + 2/5.7 +...+2/101.103
2/3 A= 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - .... - 1/101 + 1/101 - 1/103
2/3 A= 1/1 - 1/103
2/3 A= 102/103
A= 102/103 : 2/3
A= 153/103
vì 143/103 > 1 => A >1
với 1
đến đấy thì bạn tự so sánh nha!
Mình không ghi ra phân số như \(\frac{2}{3}\) nhé ( lười )
thanks all of you
A=3/1.3+3/3.5+...+3/101.103
A=3/2.(2/1.3+2/3.5+...+2/101.103)
A=3/2.(1-1/3+1/3-1/5+...+1/101-1/103)
A=3/2.(1-1/103)
A=3/2.102/103
A=153/103
Vì 153/103>1 => A>1
\(\frac{3}{3.1}=1\Rightarrow A=\frac{3}{3.1}+.......>1.\)
\(A=\frac{3}{2}\cdot\frac{3}{1\cdot3}+\frac{3}{2}\cdot\frac{3}{3\cdot5}+\frac{3}{2}\cdot\frac{3}{5\cdot7}+...+\frac{3}{2}\cdot\frac{3}{101\cdot103}\)với 1
A=\(\frac{3}{2}\)(\(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+.......+\frac{3}{101\cdot103}\))
A=\(\frac{3}{2}\)(\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\))
A=\(\frac{3}{2}\)(\(\frac{1}{1}-\frac{1}{103}\))
A=\(\frac{3}{2}\)(\(\frac{103}{103}-\frac{1}{103}\))
A=\(\frac{3}{2}\)\(\frac{102}{103}\)
A=\(\frac{306}{206}\)
\(\Rightarrow\frac{306}{206}>1\)
\(A=3\left(\frac{1}{1}-\frac{1}{3}\right)\frac{1}{3}+3\left(\frac{1}{3}-\frac{1}{5}\right)\frac{1}{2}+......+3\left(\frac{1}{101}-\frac{1}{103}\right)\frac{1}{2}\)
\(A=3x\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(A=\frac{3}{2}x\frac{102}{103}\)
\(A=\frac{153}{103}\)
\(A=1\frac{50}{103}\)
\(1\frac{50}{103}\) > 1
-> A >1
chắc chắn là A> 1 rồi
lớn hơn nha(A>1)
Mấy chế dài dòng quá, noi gương cách ngắn nhất đây này!
\(A=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{101.103}\)
\(=\frac{3}{3}+\frac{3}{3.5}+...+\frac{3}{101.103}\)
\(=1+\frac{3}{3.5}+...+\frac{3}{101.103}>\)\(1\)