Question

giai cac pt sau:

2x^2-5x+2=0

3x^2-7x-20=0

x^3+x^2+4=0

x^3-5x^2+8x-4=0

Answer 1

a) 2x²-4x-x+2=0

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

=> (2x-1)(x-2)=0

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

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

b) 3x²-12x+5x-20=0

=> 3x(x-4)+5.(x-4)=0

=> (x-4)(3x+5)=0

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x³+2x²-x²-2x+2x+4=0

=> x²(x+2)-x(x+2)+2(x+2)=0

=>(x²-x+2)(x+2)=0

=> x=-2( vi x²-x+2>0)

d) x³-x²-4x²+4x+4x-4=0

=> x²(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x²-4x+4)=0

=> \(\left[{}\begin{matrix}x-1=0\\x^2-4x+4=0\end{matrix}\right.\)

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

Answer 2

2x²-5x+2=0

⇔2x²-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

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

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x³+x²+4=0

⇔x³+2x²-x²-2x+2x+4=0

⇔(x³+2x²)-(x²+2x)+(2x+4)=0

⇔x²(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x²-x+2)=0

⇔x+2=0 và x²-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}