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a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)
\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)
\(\Leftrightarrow46x-429=0\)
\(\Leftrightarrow46x=429\)
hay \(x=\frac{429}{46}\)
Vậy: \(x=\frac{429}{46}\)
b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)
\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)
\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)
\(\Leftrightarrow-685x+261.5=0\)
\(\Leftrightarrow-685x=-261.5\)
hay \(x=\frac{523}{1370}\)
Vậy: \(x=\frac{523}{1370}\)
c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)
\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)
\(\Leftrightarrow-125x+423=0\)
\(\Leftrightarrow-125x=-423\)
hay \(x=\frac{423}{125}\)
Vậy: \(x=\frac{423}{125}\)
d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x+20x-140=0\)
\(\Leftrightarrow-37x+259=0\)
\(\Leftrightarrow-37x=-259\)
hay \(x=7\)
Vậy: x=7
8,
b, (-x2+12x+4)/(x2+3x-4) = 12/(x+4) + 12/(3x-3)
(=) (-x2+12x+4)/(x-1)(x+4) -12(x-1)/(x-1)(x+4) - 4(x+4)/(x-1)(x+4) = 0
(=) -x2 +12x +4 -12x +12 -4x -16 = 0
(=) -x2 -4x = 0
(=) -x(x+4) = 0
(=) -x = 0 hoặc x +4 = 0
(=) x=0 hoặc x=-4
Vậy S={0;4}
Chúc bạn học tốt.
a) ĐKXĐ: \(x\ne1\)
Ta có: \(\frac{7x-3}{x-1}=\frac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
hay \(x=\frac{7}{19}\)
Vậy: \(x=\frac{7}{19}\)
b) ĐKXĐ: \(x\ne-1\)
Ta có: \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=1+x\)
\(\Leftrightarrow12-28x-1-x=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow29x=11\)
hay \(x=\frac{11}{29}\)
Vậy: \(x=\frac{11}{29}\)
c) ĐKXĐ: \(x\notin\left\{\frac{-2}{3};\frac{1}{3}\right\}\)
Ta có: \(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow15x^2-8x+1-15x^2+11x+14=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
hay x=-5
Vậy: x=-5
d) ĐKXĐ: \(x\notin\left\{1;\frac{-4}{3}\right\}\)
Ta có: \(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2+7x+5=0\)
\(\Leftrightarrow44x+33=0\)
\(\Leftrightarrow44x=-33\)
hay \(x=\frac{-3}{4}\)
Vậy: \(x=\frac{-3}{4}\)
a)
\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{21x-9-2x+2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{19x-7}{3\cdot\left(x-1\right)}=0\\ \Rightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)
b)
\(\frac{2\cdot\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{12-28x}{2\cdot\left(1+x\right)}-\frac{1+x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{12-28x-1-x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{11-29x}{2\cdot\left(1+x\right)}=0\\\Rightarrow11-29x=0\\ \Rightarrow x=\frac{11}{29}\)
c)
\(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{15x^2-8x+1}{\left(3x+2\right)\cdot\left(3x-1\right)}-\frac{15x^2-11x-14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{15x^2-8x+1-15x^2+11x+14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{3x+15}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Rightarrow3x+15=0\\ \Rightarrow x=-5\)
d)
\(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\ \Leftrightarrow\frac{12x^2+37x+28}{\left(x-1\right)\cdot\left(3x+4\right)}-\frac{12x^2-7x-5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{12x^2+37x+28-12x^2+7x+5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{44x+33}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow44x+33=0\\ \Rightarrow x=-\frac{3}{4}\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
a) \(\frac{9x-0,7}{4}\)\(-\)\(\frac{5x-1,5}{7}\)=\(\frac{12x-2,1}{3}\)
⇔\(\frac{21\left(9x-0,7\right)}{84}\)\(-\)\(\frac{12\left(5x-1,5\right)}{84}\)=\(\frac{28\left(12x-2,1\right)}{84}\)
⇒189x\(-\)14,7\(-\)60x+18=336x\(-\)58,8
⇔\(-\)207x=\(-\)62,1
⇔x=\(\frac{3}{10}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{3}{10}\)}