Tìm x biết: /x+2/=/2012.x^2+4024.x/
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1:
=14,445*10=144,45
2:
=3016(2*97+1+2)=197*3016=594152
2012 x 0,86 + 0,14 x 6036 + 0,85 x 4024 + 2012 x 7,02
= 2012 x 0,86 + 0,14 x 2012 x 3 + 0,85 x 2012 x 2 + 2012 x 7,02
= 2012 x 0,86 + 0,42 x 2012 x 1,7 x 2012 + 2012 x 7,02
= 2012 x ( 0,86 + 0,42 + 1,7 + 7,02
= 2012 x 10
= 20120
Xét \( A = 1 + \dfrac{{2014}}{2} + \dfrac{{2015}}{3} + ... + \dfrac{{4023}}{{2011}} + \dfrac{{4024}}{{2012}}\\ \)
\(\Rightarrow A - 2012 = \left( {\dfrac{{2014}}{2} - 1} \right) + \left( {\dfrac{{2015}}{3} - 1} \right) + ... + \left( {\dfrac{{4024}}{{2012}} - 1} \right)\\ \Rightarrow A - 2012 = \dfrac{{2012}}{2} + \dfrac{{2012}}{3} + ... + \dfrac{{2012}}{{2012}}\\ \Rightarrow A - 2012 = 2012\left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow A = 2012\left( {1 + \dfrac{1}{2} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)503x = 2012\left( {1 + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow x = \dfrac{{2012}}{{503}} = 4 \)
\(\frac{1,4.2012.6+4,2.4024+8,4.976}{1+2+3+4+...+999}\)
\(=\frac{8,4.2012+4,2.2.2012+8,4.976}{1+2+3+4+...+999}\)
\(=\frac{8,4\left(2012+2012+976\right)}{1+2+3+4+...+999}\)
\(=\frac{8,4.5000}{500.999}\)
\(=\frac{84.500}{999.500}\)
\(=\frac{84}{999}\)
\(=\frac{28}{333}\)
1.(143 x 2012 + 30 x 4024 - 3 x 2012) : (20 : 10/11 + 23 1/7 x 7/9)
= 1 .[143 x 2012 + 60 x 2012 - 3 x 2012) : (22 + 18)
= [2012 x (143 + 60 - 3)] : 40
= [2012 x 200] : 40
= 2012 x 5
= 10060
Ủng hộ nha
1/
\(1+\frac{2014}{2}+...+\frac{4024}{2012}=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{2012}{2012}\right)\)
\(=2012+2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
Phương trình đã cho tương đương:
\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
\(\Leftrightarrow503x=2012\)
\(\Leftrightarrow x=4\)
2/
\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}+\left(1+\frac{1}{57}\right)-2-\frac{7}{3}+\frac{8}{4}=5x\)
\(\Leftrightarrow\)\(5x=\frac{17}{3}\Leftrightarrow x=\frac{17}{15}\)
3/
Ta có: \(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{n\left(n+2\right)}\right)\)\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.......\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(=2.\frac{n+1}{n+2}