Question

Решите пожалуйста!
1.(x-3)^2+(x-4)^2-(x-5)^2-x=24

2. 9x^4-37x^2=0

Answer 1

$(x-3)^{2}+(x-4)^{2}-(x-5)^{2}-x=24\(x^{2}-6x+9)+(x^{2}-8x+16)-(x^{2}-10x+25)-x=24\x^{2}-4x-x=24\x^{2}-5x-24=0\(x+3)(x-8)=0\x_{1}=-3,x_{2}=8$

Обозначаем x^2 через t:
$9t^{2}-37t+4=0\t_{1}=frac{1}{9},t_{2}=4\x^{2}=frac{1}{9},x^{2}=4\x_{1}=frac{1}{3},x_{2}=-frac{1}{3},x_{3}=2,x^{4}=-2$