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\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
a) (5x - 1)(2x + 1) = (5x -1)(x + 3)
<=> (5x - 1)(2x + 1) - (5x -1)(x + 3) = 0
<=> (5x - 1)(2x + 1 - x - 3) = 0
<=> (5x - 1)(x - 2) = 0
<=> \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,2\\x=2\end{cases}}\)
Vậy x = 0,2 ; x = 2 là nghiệm phương trình
b) x3 - 5x2 - 3x + 15 = 0
<=> x2(x - 5) - 3(x - 5) = 0
<=> (x2 - 3)(x - 5) = 0
<=> \(\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-5\right)=0\)
<=> \(x-\sqrt{3}=0\text{ hoặc }x+\sqrt{3}=0\text{ hoặc }x-5=0\)
<=> \(x=\sqrt{3}\text{hoặc }x=-\sqrt{3}\text{hoặc }x=5\)
Vậy \(x\in\left\{\sqrt{3};\sqrt{-3};5\right\}\)là giá trị cần tìm
c) (x - 3)2 - (5 - 2x)2 = 0
<=> (x - 3 + 5 - 2x)(x - 3 - 5 + 2x) = 0
<=> (-x + 2)(3x - 8) = 0
<=> \(\orbr{\begin{cases}-x+2=0\\3x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
Vậy tập nghiệm phương trình \(S=\left\{2;\frac{8}{3}\right\}\)
d) x3 + 4x2 + 4x = 0
<=> x(x2 + 4x + 4) = 0
<=> x(x + 2)2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy tập nghiệm phương trình S = \(\left\{0;-2\right\}\)
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
a. \(2x\left(x+5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2+10x-3x-2x^2=26\Leftrightarrow7x=26\Leftrightarrow x=\dfrac{26}{7}\)
Vậy \(x=\dfrac{26}{7}\)
b. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c. \(2\left(x+5\right)-x^2-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d. \(\left(2x-3\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
e. \(3x^3-48x=0\Leftrightarrow3x\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
f. \(x^3+x^2-4x=4\Leftrightarrow x^3+x^2-4x-4=0\Leftrightarrow\left(x^2-4x+4\right)+\left(x^3-8\right)=0\Leftrightarrow\left(x-2\right)^2+\left(x-2\right)\left(x^2+2x+4\right)=0\Leftrightarrow\left(x-2\right)\left(x-2+x^2+2x+4\right)=0\left(x-2\right)\left(x^2+3x+2\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+x+2x+2\right)=0\Leftrightarrow\left(x-2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=0\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=-2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)
g. \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h. \(x^2-4x+8=2x-1\Leftrightarrow x^2-4x+8-2x+1=0\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy \(x=3\)
__________________________Chúc bạn học tốt____________________________
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)