Question

Tìm x

a.(2x - 5)(3x + 4) - x(6x - 5) = 4

b.(x - 2)² + x(x - 2) = 0

c.(x³ + 4x² - x - 4) : (x + 4) = 0

d.(x⁴ + 2x³ + 10x - 25) : (x² + 5) = 0

Answer 1

a.(2x - 5)(3x + 4) - x(6x - 5) = 4

⇔ 6x² +8x -15x-20-6x²+5x=4

⇔-2x=24

⇔ x=-12

vậy x=12

b.(x - 2)² + x(x - 2) = 0

⇔(x-2)(x-2+x)=0

⇔(x-2) (2x-2)=0

⇔ (x-2)2(x-2)=0

⇔(x-2)².2=0

⇔(x-2)²=0

⇔x-2=0

⇔x=2

vậy x=2

c.(x³ + 4x² - x - 4) : (x + 4) = 0

⇔[(x³+4x²)-(x+4)] :(x+4)=0

⇔ [x²(x+4)-(x+4)] :(x+4)=0

⇔ (x+4)(x²-1):(x+4)=0

⇔(x-1)(x+1)=0

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

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