a, \(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\) Điều kiện xác định: \(x\ne-3,x\ne3\)
\(\Rightarrow\left(x-3\right)\left(2x-1\right)=\left(x+3\right)\left(2x+1\right)\)
\(\Leftrightarrow2x^2-x-6x+3=2x^2+x+6x+3\)
\(\Leftrightarrow2x^2-x-6x-2x^2-x-6x=3-3\)
\(\Leftrightarrow-14x=0\)
\(\Leftrightarrow x=0\left(TĐK\right)\)
\(\Rightarrow S=\left\{0\right\}\)
b,\(\dfrac{x^2+3}{x-2}=x+5\) Điều kiện xác định:\(x\ne2\)
\(\Rightarrow x^2+3=\left(x+5\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+3=x^2-2x+5x-10\)
\(\Leftrightarrow x^2-x^2+2x-5x=-10-3\)
\(\Leftrightarrow-3x=-13\)
\(\Leftrightarrow x=\dfrac{-13}{-3}=\dfrac{13}{3}\left(TĐK\right)\)
\(\Rightarrow S=\left\{\dfrac{13}{3}\right\}\)
c, \(2x\left(x-6\right)+3\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{-3}{2}\end{matrix}\right.\)
\(\Rightarrow S=\left\{6;\dfrac{-3}{2}\right\}\)
d, \(\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-4=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
\(\Rightarrow S=\left\{1;2;3\right\}\)
A) \(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\)
=> (2x-1)(x-3)=(x+3)(2x+1)
=> 2x2-6x-x+3=2x2+x+6x+3
=> 2x2-7x+3-2x2-7x-3=0
=>-14x=0
=> x=0
vậy S={0}
a, \(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\) (1)
ĐKXĐ của phương trình \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
pt (1)\(\Rightarrow\left(2x-1\right)\left(x-3\right)=\left(2x+1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-7x+3=2x^2+7x+3\)
\(\Leftrightarrow2x^2-2x^2-7x-7x=3-3\)
\(\Leftrightarrow-14x=0\Leftrightarrow x=0\) ( thỏa mãn )
vậy phương trình có nghiệm duy nhất x = 0
b, \(\dfrac{x^2+3}{x-2}=x+5\) (2)
ĐKXĐ : \(x-2\ne0\Leftrightarrow x\ne2\)
(2) \(\Rightarrow x^2+3=\left(x-2\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+3=x^2+3x-10\)
\(\Leftrightarrow-3x=-10-3\Leftrightarrow-3x=-13\Leftrightarrow x=\dfrac{13}{3}\)(t/m)
vây p có nghieemk duy nhất x = 13 / 3
c, \(2x\left(x-6\right)+3\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{-3}{2}\end{matrix}\right.\)
vậy tập nghiệm của pt \(S=\left\{6;\dfrac{-3}{2}\right\}\)
d, \(\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right).2\left(x-3\right).3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
vậy tập nghiệm của pt \(S=\left\{1;2;3\right\}\)
c.
\(2x\left(x-6\right)+3\left(x-6\right)=0\)
\(\Leftrightarrow2x^2-12x+3x-18=0\)
\(\Leftrightarrow2x^2-9x-18=0\)
\(\Leftrightarrow2x^2-6x-3x-18=0\)
\(\Leftrightarrow\left(x-6\right)\left(2x+3\right)=0\)
<=> x = 6 hoặc x = -3/2
