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1. \(\frac{3x-7}{5}=\frac{2x-1}{3}\)
<=> 3(3x-7)=5(2x-1)
<=> 9x-21=10x-5
<=> -21+5=10x-9x
<=> x=-16
2. \(\frac{3x-7}{2}+\frac{2x-1}{3}=-16\)
<=> \(\frac{3\left(3x-7\right)}{6}+\frac{2\left(2x-1\right)}{6}=\frac{-96}{6}\)
=> 9x-21+4x-2=-96
<=> 13x-23=-96
<=> 13x=-73
<=> x=\(\frac{-73}{13}\)
3. \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
<=> \(\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)
=> 15x-5x-5=6x+3
<=> 15x-5x-6x=3+5
<=> 4x=8
<=> x=2
4. \(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5-\left(5-2x\right)}{6}\)
<=>\(\frac{7-3x}{12}+\frac{9}{12}=\frac{24\left(x-2\right)}{12}+\frac{2\left[5-\left(5-2x\right)\right]}{12}\)
=> 7-3x+9=24x-48+4x
<=> -3x-24x-4x=-48-7
<=> -31x=-55
<=> x= \(\frac{55}{31}\)
5. \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
<=> \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+13\right)}{21}\)
=> 14x-7-15x-6=21x+273
<=> 14x-15x-21x=273+7+6
<=> -22x=286
<=> x= -13
a/\(\Leftrightarrow3\left(3x-7\right)=5\left(2x-1\right)\Leftrightarrow9x-21=10x-5\Leftrightarrow x=-16\)
b/\(\Leftrightarrow\frac{9x-21+4x-2}{6}=-16\)\(\Leftrightarrow13x-23=-96\Leftrightarrow x=x=-\frac{73}{13}\)
c/\(\Leftrightarrow\frac{3x-x+1}{3}-\frac{2x+1}{5}=0\Leftrightarrow\left(2x+1\right)\left(\frac{1}{3}-\frac{1}{5}\right)=0\Leftrightarrow x=-\frac{1}{2}\)
a/ĐKXĐ: \(y\ne4\)
Đặt \(y-4=x\)
\(1+\frac{45}{x^2}=\frac{14}{x}\Leftrightarrow x^2-14x+45=0\Rightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y-4=9\\y-4=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=13\\y=9\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne1\)
Đặt \(x-1=y\)
\(\frac{5}{y}-\frac{4}{3y^2}=3\Leftrightarrow9y^2=15y-4\)
\(\Leftrightarrow9y^2-15y+4=0\Rightarrow\left[{}\begin{matrix}y=\frac{4}{3}\\y=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{4}{3}\\x-1=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{4}{3}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne5\)
\(\Leftrightarrow2x-5=3x-15\)
\(\Leftrightarrow x=10\)
d/ ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow2\left(x^2-12\right)=2x^2+3x\)
\(\Leftrightarrow3x=-24\Rightarrow x=-8\)
e/ ĐKXĐ: \(x\ne2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=1\end{matrix}\right.\)
f/ DKXĐ: \(x\ne-\frac{1}{2}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=8\)
\(\Leftrightarrow4x^2-1=8\)
\(\Leftrightarrow x^2=\frac{9}{4}\Rightarrow x=\pm\frac{3}{2}\)
\(b.\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(dkxd:x\ne\pm2\right)\\ \Leftrightarrow\frac{12}{x^2-4}-\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\frac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\\\Leftrightarrow 12-x^2-3x-2+x^2+5x-14=0\\ \Leftrightarrow2x-4=0\\\Leftrightarrow 2\left(x-2\right)=0\\\Leftrightarrow x-2=0\\\Leftrightarrow x=2\left(ktmdk\right)\)
Vô nghiệm
\(a.\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\left(dkxd:x\ne\pm1\right)\\\Leftrightarrow \frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\\\Leftrightarrow \left(x+1\right)^2-\left(x-1\right)^2=16\\\Leftrightarrow \left(x+1-x+1\right)\left(x+1+x-1\right)-16=0\\\Leftrightarrow 4x-16=0\\\Leftrightarrow 4\left(x-4\right)=0\\\Leftrightarrow x-4=0\\ \Leftrightarrow x=4\left(tmdk\right)\)
a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow-21x=3x-60\)
\(\Leftrightarrow24x=60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{\left(8x-3\right)-2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)+\left(x+3\right)}{4}\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow x=-16\)
Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)
\(\Leftrightarrow27-9x+80-16x=12-12x-48\)
\(\Leftrightarrow-25x+107=-12x-36\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
a, \(2x-\frac{1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{1}{2}\left(4x-1\right)=\frac{1}{8}\left(6x+1\right)\)
\(\Leftrightarrow4\left(4x-1\right)=6x+1\)
\(\Leftrightarrow10x=5\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b, \(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x = 3
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=0+3\)
\(\Leftrightarrow x=3\)