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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}
Ta có: \(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+...+\frac{8}{49.57}\right)+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
\(\Leftrightarrow1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+\frac{1}{17}-\frac{1}{25}+....+\frac{1}{49}-\frac{1}{57}+2x-2=\frac{8x+28+15x-24}{12}\)
\(\Leftrightarrow1-\frac{1}{57}+2x-2=\frac{23x+4}{12}\)
\(\Leftrightarrow2x-\frac{58}{57}=\frac{23x+4}{12}\)
\(\Leftrightarrow24x-\frac{232}{19}=23x+4\)
\(\Leftrightarrow x=\frac{308}{19}\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(ĐKXĐ:x\ne1;-1;2;-2\)
\(\frac{\left(x+4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-8\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+8\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{8}{3}\)
\(\Leftrightarrow\frac{x^2+x+4x+4+x^2-x-4x+4}{x^2-1}=\frac{x^2-2x-8x+16+x^2+2x+8x+16}{x^2-4}-\frac{8}{3}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{2x^2+32}{x^2-4}-\frac{8}{3}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{3\left(2x^2+32\right)}{3\left(x^2-4\right)}-\frac{8\left(x^2-4\right)}{3\left(x^2-4\right)}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{9x^2+96-8x^2+32}{3\left(x^2-4\right)}\)
\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{x^2+128}{3\left(x^2-4\right)}\)
\(\Leftrightarrow3\left(x^2-4\right)\left(2x^2+8\right)=\left(x^2+128\right)\left(x^2-1\right)\)
\(\Leftrightarrow9x^4+24x^2-24x^2-96=x^4-x^2+128x^2-128\)
\(\Leftrightarrow9x^4+24x^2-24x^2-x^4+x^2+128x^2=-128+96\)
\(\Leftrightarrow8x^4+129x^2=-32\)
\(\Leftrightarrow8x^4+129x^2+32=0\)
\(\Leftrightarrow x=\frac{1}{2}\left(tmđkxđ\right)\)
bạn sai r bạn ơi cái chỗ chuyển vế dòng tương đương số 8 : x^4 - x^2 + 128x^2 - 128 đáng ra sau khi chuyển px là -x^4 +x^2 - 128x^2 + 128 chứ sao lại là x^4 + x^2 + 128x^2 +128
Ta có: \(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}=1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}=1-\frac{1}{57}=\frac{56}{57}\)
Vậy: 56/57 + 58/57 + 2x - 2 = 2x + 7/3 + 5x - 8/4
2 + 2x - 2 = 2x + 7/3 + 5x - 8/4
2x = 2x + 7/3 + 5x - 8/4
=> 7/3 + 5x - 8/4 = 0
1/3 + 5x = 0
=> 5x = -1/3
=> x = -1/3 : 5=-1/15
a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)
\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)
\(\Leftrightarrow96x+744=-6x+48\)
\(\Leftrightarrow102x=-696\)
\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)
Vậy .....
b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)
\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)
\(\Leftrightarrow x=-5\) (nhận)
Vậy ....
