`(2/1.2 + 2/3.4 + ... + 2/99.100) . (x^2 +x+1945)/2 > 1975 . (1/51 + 1/52 + ... + 1/99 + 1/100)`
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giải bt:tìm x, biết:2x/1.2+2x/3.4+...+2x/99.100=1/51+1/52+...+1/100
\(\Leftrightarrow2x\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100};B=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Leftrightarrow A=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{50}\\ \Leftrightarrow A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Leftrightarrow2x.A=B\Leftrightarrow2x.B-B=0\\ \Leftrightarrow B\left(2x-1\right)=0\\ \Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
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Tìm x, biết: (1/1.2+1/3.4+...+1/99.100).x=2012/51+2012/52+...+2012/99+2012/100
giúp mình nhaaaaaaaaaaaaaaaaaaa ^^ (cho bạn luông 1 0 tick í í í ^.*)
Ta có:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> x = 2012
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Giải bất phương trình:
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right).\left(x^2-x+1982\right)< 2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
tim x biet:
: \(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
Xét vế trái biểu thức, ta có:
\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot x\)
\(=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\right]\cdot x\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x\)
Xét vế phải biểu thức, ta có:
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
Từ đầu bài và 2 kết luận trên, ta suy ra:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
\(\Rightarrow x=2012\)
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A=1/1.2+1/3.4+1/5.6+....+1/97.98+1/99.100 B=1/50+1/51+1/52+....+1/99+1/100 Tính A-B
Tìm x:
\(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).2013x=2012\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
a) dfrac{2}{1^2}.dfrac{6}{2^2}.dfrac{12}{3^2}.dfrac{20}{4^2}.dfrac{30}{5^2}.....dfrac{110}{10^2}.x-20
b) left(1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{2013}right).x+2013dfrac{2014}{1}+dfrac{2015}{2}+...+dfrac{4025}{2012}+dfrac{4026}{2013}
c) left(dfrac{1}{1.2}+dfrac{1}{3.4}+...+dfrac{1}{99.100}right).xdfrac{2012}{51}+dfrac{2012}{52}+...+dfrac{2012}{99}+dfrac{2012}{100}
Đọc tiếp
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
CHO: A=1/51+1/52+....+1/100
B=1-1/2+1/3-1/4+...+1/99-1/100
C=1/1.2+1/3.4+1/5.6+...+1/99.100
SO SÁNH A VỚI B VỚI C.
CÁC BẠN GIÚP MÌNH NHA!
cái này anh mình biêt đấy bạn ơi
Ta có : B=1-1/2+1/3-...+1/99-1/100= ( 1+1/3+...+1/99) -(1/2+....+1/100)= ( 1+1/2+1/3+....+1/99+1/100)-2.(1/2+...+1/100) =1+1/2+1/3+...+1/100 - ( 1+...+1/50) = (1+1/2+...+1/50) + ( 1/51+1/52+...+1/100) - ( 1+...+1/50)= 1/51 +1/52+...+1/100 (1)
C=1/(1.2) +1/(3.4) +...+1/(99.100) = 1-1/2+ 1/3-1/4+...+1/99-1/100 =...
Biểu thức C phần còn lại làm tương tự giống phần (1) nhé => C= 1/51+1/52+...+1/100 (2)
A=1/51+...+1/100(3)
Từ (1),(2) và (3)=>A=B=C (đpcm) . Chúc cậu học tốt !
KẾT BẠN VỚI MÌNH NHA
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
CMR:
1, A = \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
2, \(\frac{25}{75}+\frac{25}{100}< A< \frac{25}{51}+\frac{25}{75}\)
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