Question

Пожалуйста, решите 3 и 7, даю 100 баллов

Answer 1

Ответ:

Объяснение:

3.

$\left \{ {{3x+5y=8} \atop {-3x+y=-2}} \right.(+) = > 6y=6= > y=1\\3x+5=8= > 3x=3= > x=1\\(1;1)$

7.

$x^{2} +\frac{16}{x^{2} }-5x+\frac{20}{x}=2\\\\\frac{x^4+16-5x^3+20x-2x^2}{x^2}=0\\ x^4-5x^3-2x^2+20x+16=0\\x^4+x^3-6x^3-6x^2+4x^2+4x+16x+16=0\\x^3(x+1)-6x^2(x+1)+4x(+1)+16(x+1)=0\\(x+1)(x^3-6x^2+4x+16)=0\\(x+1)(x^3-4x^2-2x^2+8x-4x+16)=0\\(x+1)(x^2(x-4)-2x(x-4)-4(x-4))=0\\(x+1)(x-4)(x^2-2x-4)=0\\x_1=-1 x_2=4 \\x^2-2x-4=0\\D=k^2-ac=1+4=5\\x_3=1-\sqrt{5}\\x_4=1+\sqrt{5}$