\(\frac{4}{2 . 4} + \frac{4}{4 . 6} + \frac{4}{6 . 8} + ... + \frac{4}{2012 . 2014} + \frac{4}{2014 . 2016}\) = ?
\frac{x-10}{2010}+\frac{x-8}{2012}+\frac{x-6}{2014}+\frac{x-4}{2016}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2016}{4}+\frac{x-2014}{6}+\frac{x-2012}{8}+\frac{x-2010}{10}
\(B=\frac{1-3}{1\cdot3}+\frac{2-4}{2\cdot4}+\frac{3-5}{3\cdot5}+\frac{4-6}{4\cdot6}+............+\frac{2011-2013}{2011.2013}+\frac{2012-2014}{2012\cdot2014}-\frac{2013+2014}{2013\cdot2014}\)
giải phương trình:
a)\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
b)\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)
\(\Rightarrow x+10=0\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)
\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0
Vậy x = 2012
a, (x+1)/9 +1 + (x+2)/8 = (x+3)/7 + 1 + (x+4)/6 + 1
<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6
<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0
vì 1/9 +1/8 -1/7 - 1/6 khác 0
=> x+10=0
=> x=-10
\(\text{Tính : }\)
\(B=2-4-6+8+10-12-14+16+...+2010-2012-2014+2016\)
\(C=\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{990}\)
B = 2 - 4 - 6 + 8 + 10 - 12 -14 + 16 + ...+ 2010 - 2012 - 2014 + 2016
= (2 - 4 - 6 + 8 ) + ( 10 - 12 -14 +16 ) + ...+ ( 2010 - 2012 - 2014 + 2016 )
= 0 + 0 +...+ 0 + 0 (có 252 số hạng 0)
= 0
Ta có: Từ 2 đến 2016 có \(\frac{2016-2}{2}+1=1008\)
B=(2-4)-(6-8)+(10-12)-(14-16)+...+(2010-2012)-(2014-2016) như vậy có 1008:2=504 nhóm
=-2+2-2+2+...-2+2 có 504 số hạng trong đó có: 525 số -2 và 525 số +2
=0
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{9.10.11}\)
\(\Rightarrow2.C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{9.10.11}=\frac{1}{2}\left(\frac{2}{1.3}\right)+\frac{1}{3}\left(\frac{2}{2.4}\right)+\frac{1}{4}\left(\frac{2}{3.5}\right)+\)\(...+\frac{1}{10}\left(\frac{2}{9.11}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{4}\left(\frac{1}{3}-\frac{1}{5}\right)+...+\frac{1}{10}\left(\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{3.2}-\frac{1}{3.4}+\frac{1}{4.3}-\frac{1}{4.5}+...+\frac{1}{10.9}-\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{10.11}=\frac{27}{55}\)
=> C=27/110
1.Tìm tất cả các số tự nhiên n thỏa mãn:
\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right)^{2n -1}+n.2^n=8192\)
2. So sánh A và B biết:
\(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{5.6}+...+\frac{2011}{1999.2000}\)
\(B=\frac{2012}{1001}+\frac{2012}{1002}+\frac{2012}{1003}+...+\frac{2012}{2000}\)
3. Tính \(\left(S-P\right)^{2016}\) biết:\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
4.Tìm x:
a) \(-1\frac{1}{56}:\left(\frac{1}{8}-\frac{1}{7}\right)-\frac{22}{\left|2.x-0,5\right|}=-1\frac{1}{30}:\left(\frac{1}{5}-\frac{1}{6}\right)\)
b) \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}....\frac{30}{62}.\frac{31}{64}=2^x\)
c) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
A =\(\frac{2015+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+\frac{2011}{5}+.....+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2016}}=\)
tìm A
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
tim x
\(\frac{x+1}{2016}+\frac{x+2}{2015}+\frac{x+3}{2014}=\frac{x+4}{2013}+\frac{x+5}{2012}+\frac{x+6}{2011}\)
cộng 1 vào mỗi tỉ số ta được:
\(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1+\frac{x+3}{2014}+1=\frac{x+4}{2013}+1+\frac{x+5}{2012}+\frac{x+6}{2011}\)
=>\(\frac{x+1}{2016}+\frac{2016}{2016}+\frac{x+2}{2015}+\frac{2015}{2015}+\frac{x+3}{2014}+\frac{2014}{2014}=\frac{x+4}{2013}+\frac{2013}{2013}+\frac{x+5}{2012}+\frac{2012}{2012}+\frac{x+6}{2011}+\frac{2011}{2011}\)
=>
\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}=\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\)
=>
\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\left(\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\right)=0\)
=>
\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
=>(x+2017).(1/1016+1/2015+1/2014-1/2013-1/2012-1/2011)=0
dễ thấy 1/2016<1/2015<1/2014<1/2013<1/2012<1/2011
=>1/2016+...-1/2011 khác 0
=>x+2017=0
=>x=-2017
nhớ tick
Cho A= \(\frac{4+\frac{4}{2012}-\frac{4}{2013}+\frac{4}{2014}-\frac{4}{2015}}{\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}+7}\)
Và B= \(\frac{1+2+2^2+...+2^{2013}}{2^{2015}-2}\)
Tính A - B
p/S: LM ƠN GIÚP TỚ VS :
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
ĐẶT \(C=1+2+...+2^{2013}\)
\(\Rightarrow2C=2+2^2+...+2^{2014}\)
\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)
\(\Rightarrow C=2^{2014}-2\)
\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)
\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)
\(B=\frac{1}{2}\)
\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)
\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)
VẬY, \(A-B=\frac{-1}{14}\)
Tính nhanh
\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
Giải tự luận hộ mình nha!!!!!!!! Mình cảm ơn!!!
Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
UwU
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đùa thôi đáp án: 2015 nha bn
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