Giải phương trình:  4(x + 5)(x + 6)(x + 10)(x + 12) = 3x².

Ta có 4(x + 5)(x + 6)(x + 10)(x + 12) = 3x²

⇔ 4(x + 5)(x + 6)(x + 10)(x + 12) – 3x² = 0

⇔ 4[(x + 5)(x + 12)][(x + 10)(x + 6)] – 3x² = 0

⇔ 4(x² + 17x + 60)(x² + 16x + 60) – 3x² = 0                    (*)

Đặt x² + 16x + 60 = a

Suy ra x² + 17x + 60 = a + x

Khi đó (*) ⇔ 4a(a + x) – 3x² = 0

⇔ 4a² + 4ax – 3x² = 0

⇔ 4a² + 4ax + x² – 4x² = 0

⇔ (2a + x)² – 4x² = 0

⇔ (2a + x – 2x)(2a + x + 2x) = 0

⇔ (2a – x)(2a + 3x) = 0

⇔ [2(x² + 16x + 60) – x][2(x² + 16x + 60) + 3x] = 0

⇔ (2x² + 32x + 120 – x)(2x² + 32x + 120 + 3x) = 0

⇔ (2x² + 31x + 120)(2x² + 35x + 120) = 0

⇔ (2x² + 16x + 15x + 120)(2x² + 35x + 120) = 0

⇔ (2x + 15)(x + 8)(2x² + 35x + 120) = 0

$\iff \left[\begin{array}{l}2\text{x}+15=0\\ x+8=0\\ 2{\text{x}}^2+35\text{x}+120=0\end{array}\right.$

$\iff \left[\begin{array}{l}\text{x=}\frac{-15}{2}\\ x=-8\\ x=\frac{-35+\sqrt{265}}{4}\\ x=\frac{-35-\sqrt{265}}{4}\end{array}\right.$

Vậy $x\in \left\{\frac{-15}{2};-8;\frac{-35+\sqrt{265}}{4};\frac{-35-\sqrt{265}}{4}\right\}$  .