Question

câu 2) tìm x biết 

a) x ( 3 - 2 x ) + 2x² = 12

b) x(x - 3 ) - x + 3 = 0

c) x³ - 3x²  + 3x - 1 = 0

d) ( 3x - 1 )² - ( x² - 2x + 1 ) = 0

Answer 1

a: \(x\left(3-2x\right)+2x^2=12\)

=>\(3x-2x^2+2x^2=12\)

=>3x=12

=>\(x=\dfrac{12}{3}=4\)

b: x(x-3)-x+3=0

=>x(x-3)-(x-3)=0

=>(x-3)(x-1)=0

=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c: \(x^3-3x^2+3x-1=0\)

=>\(x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)

=>\(\left(x-1\right)^3=0\)

=>x-1=0

=>x=1

d: \(\left(3x-1\right)^2-\left(x^2-2x+1\right)=0\)

=>\(9x^2-6x+1-x^2+2x-1=0\)

=>\(8x^2-4x=0\)

=>\(2x^2-x=0\)

=>x(2x-1)=0

=>\(\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)