Question

Giải pt: (x²+x-6)(x²+9x+14) = 300

(2x-5)(3x+1) = 4x²-25

Answer 1

a) (x²+x-6)(x²+9x+14) = 300

<=> (x-2)(x+3)(x+2)(x+7) - 300 = 0

<=> [(x-2)(x+7)][(x+2)(x+3)] - 300 = 0

<=> (x²-5x-14)(x²+5x+6) - 300 = 0

Đặt x² + 5x - 14 = a

<=> a(a+20) - 300 = 0

<=> a² + 20a - 300 = 0

<=> a² + 20a + 100 - 400 = 0

<=> (a+10)² - 20² = 0

<=> (a-10)(a+30) = 0

<=> \(\left[{}\begin{matrix}a=10\\a=-30\end{matrix}\right.\)

Với a = 10, ta có:

x² + 5x - 14 = 10

=> x² + 5x - 24 = 0

=> (x-3)(x+8) = 0

=> \(\left[{}\begin{matrix}x=3\\x=-8\end{matrix}\right.\)

Với a = -30, ta có:

x² + 5x - 14 = -30

=> x² + 5x + 16 = 0 (vn)

Vậy nghiệm pt x = 3; x = -8

b) (2x-5)(3x+1) = 4x² - 25

<=> (2x-5)(3x+1) = (2x-5)(2x+5)

<=> (2x-5)(3x+1-2x-5) = 0

<=> (2x-5)(x-4) = 0

<=> \(\left[{}\begin{matrix}2x-5=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=4\end{matrix}\right.\)

Vậy...