a. \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\)
\(\Leftrightarrow-x^3+5x^2-5x=0\)
\(\Leftrightarrow-x\left(x^2-5x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)
a) \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-2-x^2+4x-3\right)=0\)
\(\Leftrightarrow x\left(-x^2+5x-5\right)=0\)
\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)
=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)
b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)
\(\Leftrightarrow-12=0\left(vn\right)\)
c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)
\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)
\(\Leftrightarrow2x=15\)
\(\Rightarrow x=\frac{15}{2}\)
a) x( x - 2 ) - x( x - 1 )( x - 3 ) = 0
<=> x2 - 2x - x( x2 - 4x + 3 ) = 0
<=> x2 - 2x - x3 + 4x2 - 3x = 0
<=> -x3 + 5x2 - 5x = 0
<=> -x( x2 - 5x + 5 ) = 0
<=> \(\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\)
+) -x = 0 => x = 0
+) x2 - 5x + 5 = 0 (*)
\(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot1\cdot5==25-20=5\)
\(\Delta>0\)nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{5+\sqrt{5}}{2}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{5-\sqrt{5}}{2}\end{cases}}\)
Vậy S = { \(0;\frac{5\pm\sqrt{5}}{2}\)}
b) ( 2x - 5 )( x + 3 ) - ( x - 1 )( 2x + 3 ) = 0
<=> 2x2 + x - 15 - ( 2x2 + x - 3 ) = 0
<=> 2x2 + x - 15 - 2x2 - x + 3 = 0
<=> -12 = 0 ( vô lí )
Vậy phương trình vô nghiệm
c) ( x - 2 )( x2 + 2x + 8 ) - x3 - 2x + 1 = 0
<=> x3 + 4x - 16 - x3 - 2x + 1 = 0
<=> 2x - 15 = 0
<=> 2x = 15
<=> x = 15/2
a, \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\Leftrightarrow x^2-2x-x\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\Leftrightarrow5x^2-5x-x^3=0\) ( vô nghiệm )
b, \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\Leftrightarrow2x^2+6x-5x-15-\left(2x^2+x+3\right)=0\)
\(\Leftrightarrow2x^2+x-15-2x^2-x-3=0\Leftrightarrow-12\ne0\)( vô nghiệm )
c, \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)
\(\Leftrightarrow x^3+2x^2+8x-2x^2-4x-16-x^3-2x+1=0\Leftrightarrow2x-15=0\Leftrightarrow x=\frac{15}{2}\)
