19. ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
PT \(\Leftrightarrow\dfrac{2-3x-1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow1-3x=\dfrac{1}{x-2}\)
\(\Leftrightarrow\left(1-3x\right)\left(x-2\right)=1\)
\(\Leftrightarrow x-3x^2-2+6x-1=0\)
\(\Leftrightarrow3x^2-7x+3=0\)
\(\Leftrightarrow x=\dfrac{7\pm\sqrt{13}}{6}\) ( TM )
Vậy ...
Câu 20 :
ĐKXĐ : \(x\ne2\)
PT \(\Leftrightarrow\dfrac{2\left(x+5\right)}{6x-12}-\dfrac{3x-6}{6x-12}=\dfrac{3\left(2x-3\right)}{6x-12}\)
\(\Leftrightarrow2\left(x+5\right)-\left(3x-6\right)=3\left(2x-3\right)\)
\(\Leftrightarrow2x+10-3x+6=6x-9\)
\(\Leftrightarrow2x+10-3x+6-6x+9=0\)
\(\Leftrightarrow-7x+25=0\)
\(\Leftrightarrow x=\dfrac{25}{7}\) ( TM )
Vậy ...
19: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
Ta có: \(\dfrac{2}{x+1}-\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{2-3x-1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{-3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(-3x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
Suy ra: \(-3x^2+6x+x-2-1=0\)
\(\Leftrightarrow-3x^2+7x-3=0\)
\(\Delta=7^2-4\cdot\left(-3\right)\cdot\left(-3\right)=13\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là
\(\left\{{}\begin{matrix}x_1=\dfrac{-b-\sqrt{\Delta}}{2a}\\x_2=\dfrac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-7-\sqrt{13}}{2\cdot\left(-3\right)}=\dfrac{7+\sqrt{13}}{6}\left(nhận\right)\\x_2=\dfrac{-7+\sqrt{13}}{2\cdot\left(-3\right)}=\dfrac{7-\sqrt{13}}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{7+\sqrt{13}}{6};\dfrac{7-\sqrt{13}}{6}\right\}\)
2) ĐKXĐ: \(x\ne2\)
Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)
Suy ra: \(2x+10-3x+6=6x-9\)
\(\Leftrightarrow-x+16-6x+9=0\)
\(\Leftrightarrow-7x+25=0\)
\(\Leftrightarrow-7x=-25\)
hay \(x=\dfrac{25}{7}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{25}{7}\right\}\)
19) ĐKXĐ: \(x\ne-1;x\ne2\)
\(\dfrac{2}{x+1}-\dfrac{3x+1}{\left(x+1\right)}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{2}{x+1}-\dfrac{3x+1}{\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)-\left(3x+1\right)\left(x-2\right)-1}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow-3x^2+7x-3=0\)
\(\Leftrightarrow-3\left(x^2-\dfrac{7}{3}+\dfrac{49}{36}\right)=-\dfrac{13}{12}\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{13}{36}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7+\sqrt{13}}{6}\left(tm\right)\\x=\dfrac{7-\sqrt{13}}{6}\left(tm\right)\end{matrix}\right.\)
Vậy tập nghiệp của PT trên là: \(S=\left\{\dfrac{7+\sqrt{13}}{6};\dfrac{7-\sqrt{13}}{6}\right\}\)
20) ĐKXĐ: \(x\ne2\)
\(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{x+5}{3\left(x-2\right)}-\dfrac{1}{2}-\dfrac{2x-3}{2\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x+10-3x+6-6x+9}{6\left(x-2\right)}=0\)
\(\Leftrightarrow-7x+25=0\Leftrightarrow x=\dfrac{25}{7}\)
Vậy tập nghiệm của PT trên là \(S=\left\{\dfrac{25}{7}\right\}\)