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\(\frac{2}{x-14}\) -\(\frac{5}{x-13}\)= \(\frac{2}{x-9}\)- \(\frac{5}{x-11}\)
ĐKXĐ : x\(\ne\)14;13;9;11
(=) \(\frac{2}{x-14}\)- \(\frac{5}{x-13}\)= \(\frac{2}{x-9}\)-\(\frac{5}{x-11}\)
(=) \(\frac{2}{x-14}\)-\(\frac{2}{x-9}\)=\(\frac{5}{x-13}\)-\(\frac{5}{x-11}\)
(=) \(\frac{2\left(x-9\right)}{\left(x-14\right)\left(x-9\right)}\)-\(\frac{2\left(x-14\right)}{\left(x-14\right)\left(x-9\right)}\)=\(\frac{5\left(x-11\right)}{\left(x-13\right)\left(x-11\right)}\)-\(\frac{5\left(x-13\right)}{\left(x-13\right)\left(x-11\right)}\)
(=) \(\frac{2x-18-2x+28}{\left(x-14\right)\left(x-9\right)}\)=\(\frac{5x-55-5x+65}{\left(x-13\right)\left(x-11\right)}\)
(=) \(\frac{10}{\left(x-14\right)\left(x-9\right)}\)=\(\frac{10}{\left(x-13\right)\left(x-11\right)}\)
=) ( x - 14 ) ( x - 9 ) = ( x - 13 ) ( x - 11 )
(=) x2 - 9x - 14x + 126 = x2 - 13x - 11x + 143
(=) x - 17 = 0
(=) x = 17
Vậy phương trình có nghiệm là: x = 17
2.
\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
\(\Rightarrow x>0\)
Vậy \(x>0\)
(x-5)(x-9)>0\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\Leftrightarrow x>5\\x-9>0\Leftrightarrow x>9\end{matrix}\right.\)
Vậy x>9 thì (x-5)(x-9)>0
a.
\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)
\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=28-4x\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)
\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=-4x+28\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
Vậy ................................
bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
ĐKXĐ \(x\ne8;x\ne11;x\ne9;x\ne10\)
\(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)
\(\Leftrightarrow\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)
\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}=\dfrac{x}{x-9}+\dfrac{x}{x-10}\)
\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)
1) x=0
2) \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)
\(\Leftrightarrow\dfrac{x-11+x-8}{\left(x-8\right)\left(x-11\right)}-\dfrac{x-10+x-9}{\left(x-9\right)\left(x-10\right)}=0\)
\(\Leftrightarrow\dfrac{2x-19}{\left(x-8\right)\left(x-11\right)}=\dfrac{2x-19}{\left(x-9\right)\left(x-10\right)}\)
\(\Leftrightarrow\dfrac{2x-19}{x^2-19x+88}=\dfrac{2x-19}{x^2-19x+90}\)
do \(x^2-19x+88\ne x^2-19x+90\)
\(\Rightarrow2x-19=0\)
=> x=\(\dfrac{19}{2}\)
Vậy x=\(0\); x=\(\dfrac{19}{2}\)
Tik 
a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
\(\dfrac{2}{x-14}-\dfrac{5}{x-13}=\dfrac{2}{x-9}-\dfrac{5}{x-11}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne9;11;13;14\\\left(\dfrac{2}{x-14}-\dfrac{2}{3}\right)-\left(\dfrac{5}{x-13}-\dfrac{5}{4}\right)=\left(\dfrac{2}{x-9}-\dfrac{1}{4}\right)-\left(\dfrac{5}{x-11}-\dfrac{5}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow2\left(\dfrac{x-17}{3\left(x-14\right)}\right)-5\left(\dfrac{x-17}{4\left(x-13\right)}\right)=\left(\dfrac{x-17}{4\left(x-9\right)}\right)-5\left(\dfrac{x-17}{6\left(x-11\right)}\right)\)
\(\left(x-17\right)\left[\dfrac{2}{3\left(x-14\right)}-\dfrac{5}{4\left(x-13\right)}+\dfrac{5}{6\left(x-11\right)}-\dfrac{1}{4\left(x-9\right)}\right]=0\)
[..] vô nghiệm
x=17
Lời giải:
Bài của bạn ngonhuminh cơ bản không đúng do không có cơ sở khẳng định biểu thức trong ngoặc vuông vô nghiệm.
ĐKXĐ: \(x\neq \left\{9;11;13;14\right\}\)
\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)
\(\Leftrightarrow 2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)=5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)\)
\(\Leftrightarrow \frac{10}{(x-14)(x-9)}=\frac{10}{(x-13)(x-11)}\)
\(\Rightarrow (x-14)(x-9)=(x-13)(x-11)\)
\(\Leftrightarrow x^2-23x+126=x^2-24x+143\)
\(\Leftrightarrow x-17=0\Leftrightarrow x=17\)
Thử lại thấy thỏa mãn.
Vậy \(x=17\)