Question

x2+8x−72/x+81/x2=18

20 балов срочно

Answer 1

Ответ:

X1=-9,X2=-3,X3=1,X4=3

Объяснение:

X2+8x-72/x+81/x2=18

x2+8x-72/x+81/x2=18,$x\neq 0$

$x^{2} +8x-\frac{72}{x} +\frac{81}{x^{2} } -18=0$

$\frac{x^{4}+8x^{3}-72x+81-18x^{2} }{x^{2} } =0$

$x^{4} +8x^{3} -18x^{2} -72x+81=0$

$x^{3} *(x-1)+9x^{2} *(x-1)-9x*(x-1)-81(x-1)=0$

$(x-1)*(x^{3} +9x^{2} -9x-81)=0$

$(x-1)*(x^{2} *(x+9)-9(x+9))=0$

$(x-1)*(x+9)*(x^{2} -9)=0$

$x-1=0\\x+9=0\\x^{2} -9=0$

$x=1\\ x=-9\\x=-3\\x=3$         x$\neq 0$

ответ:

$x1=-9\\x2=-3\\x3=1\\x4=3$