Question

ДАЮ 24 БАЛЛА (X^2/(X^2-7X+10))+(16/(3X^2-12))=1

Answer 1

(X^2/(X^2-7X+10))+(16/(3X^2-12))=1
$\frac{x^2}{x^2-7x+10} + \frac{16}{3x^2-12} =1 \\ \\ \frac{x^2}{(x-5)(x-2)} + \frac{16}{3(x^2-4)} -1=0 \\ \\ \frac{x^2}{(x-5)(x-2)} + \frac{16}{3(x-2)(x+2)} -1=0 \\ \\ \frac{3x^2(x+2)+16(x-5)-3(x-5)(x-2)(x+2)}{3(x-5)(x-2)(x+2)} =0 \\ \\ \frac{3x^3+6x^2+16x-80-(3x-15)(x^2-4)}{3(x-5)(x-2)(x+2)} =0 \\ \\ \frac{3x^3+6x^2+16x-80-3x^3+15x^2+12x-60}{3(x-5)(x-2)(x+2)} =0 \\ \\ \frac{21x^2+28x-140}{3(x-5)(x-2)(x+2)} =0 \\ \\ \frac{3x^2+4x-20}{(x-5)(x-2)(x+2)} =0$

ОДЗ : x≠5; x≠2; x≠-2
3x² + 4x - 20 = 0
D/4 = 2² + 3 * 20 = 64 = 8²
x₁ = (-2 + 8) / 3 = 2    - не подходит по ОДЗ
x₂ = (-2 - 8) / 3 = $- \frac{10}{3} = -3 \frac{1}{3}$

Ответ: х = $-3 \frac{1}{3}$