Question

Tìm x:

a) x(x - 1) + (x + 2)^2 = 1

b) (x + 5).(x + 3) - (x-2)^2 = -1

c) x(2x - 2) - (x-2)(2x + 3) = 0

d) 3x^3 - 27x = 0

GIúp mình với! Cần gấp lắm ạ!!!

Answer 1

a) x(x-1) + (x - 2)² = 1

<=> x² -x + x² - 4x + 4 =1

<=>2x² -5x + 3 =0

<=> 2x² - 2x - 3x +3 =0

<=> 2x(x-1) - 3(x-1)=0

<=> (x - 1)(2x - 3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy

b) (x + 5)(x + 3) - (x - 2)² = -1

<=> x² + 3x + 5x + 15 - x² +4x - 4 = -1

<=>12x + 12= 0

<=> 12(x - 1)=0

<=> x -1 = 0

<=.x=1

Vậy

c) x(2x-2) - (x-2)(2x+3)=0

<=>2x² - 2x - 2x² -3x + 4x +6 =0

<=> -x + 6 = 0

<=> x=6

vậy

d) 3x³ - 27x =0

<=> 3x(x² - 9)=0

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

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

vậy