a, \(\dfrac{5}{4x+5}=\dfrac{-2}{3x-1}\) (ĐKXĐ: \(x\ne-\dfrac{5}{4};x\ne\dfrac{1}{3}\))
\(\Leftrightarrow\dfrac{5\left(3x-1\right)}{\left(4x+5\right)\left(3x-1\right)}=\dfrac{-2\left(4x+5\right)}{\left(4x+5\right)\left(3x-1\right)}\)
\(\Rightarrow15x-5=-8x-10\)
\(\Leftrightarrow15x+8x=-10+5\)
\(\Leftrightarrow23x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{23}\) (Thỏa mãn)
Vậy pt có tập nghiệm là \(S=\left\{-\dfrac{5}{23}\right\}\)
b, \(\dfrac{2x}{x+2}=\dfrac{2x+3}{x-11}\) (ĐKXĐ: \(x\ne-2;x\ne11\))
\(\Leftrightarrow\dfrac{2x\left(x-11\right)}{\left(x+2\right)\left(x-11\right)}=\dfrac{\left(2x+3\right)\left(x+2\right)}{\left(x+2\right)\left(x-11\right)}\)
\(\Rightarrow2x^2-22x=2x^2+7x+6\)
\(\Leftrightarrow2x^2-22x-2x^2-7x=6\)
\(\Leftrightarrow-29x=6\)
\(\Leftrightarrow x=-\dfrac{6}{29}\) (Thỏa mãn)
Vậy pt có tập nghiệm là \(S=\left\{-\dfrac{6}{29}\right\}\)
c, \(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\) (ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{1-x}{x+1}+\dfrac{3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)
\(\Rightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow-x+3x+3-2x-3=-1\)
\(\Leftrightarrow0x=-1\) (vô lý)
Vậy pt vô nghiệm
d, \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\) (ĐKXĐ: \(x\ne\pm2\))
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(3x-2\right)+1}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow-6x^2-11x+2+9x^2-14x-8=3x^2-2x+1\)
\(\Leftrightarrow-6x^2-11x+2+9x^2-14x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\) (thỏa mãn)
Vậy pt có tập nghiệm là \(S=\left\{-\dfrac{7}{23}\right\}\)
e, \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\) (ĐKXĐ: \(x\ne2;x\ne4\))
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Rightarrow x^2-7x+12+x^2-4x+4=-x^2+6x-8\)
\(\Leftrightarrow x^2-7x+12+x^2-4x+4+x^2-6x+8=0\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\) (Thỏa mãn)
Vậy pt có tập nghiệm là \(S=\left\{\dfrac{8}{3};3\right\}\)
f, \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3;x\ne\dfrac{7}{2}\))
\(\Leftrightarrow\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
\(\Rightarrow13x+39+x^2-9=12x+42\)
\(\Leftrightarrow13x+39+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(KTM\right)\\x=-4\left(TM\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm là \(S=\left\{-4\right\}\)
