Question

Решить систему уравнений №9

Answer 1

Решение во вложении:

Answer 2

$left{begin{array}{l}(sqrt3)^{3(frac{x}{y}+frac{y}{x})}=243\log_2(y+x)+log_2(y-x)=3end{array}right; ; ,; ; ; ; ODZ:; ; left { {{xne 0; yne 0} atop {y>-x; ,; y>x}} right. \\\left{begin{array}{ll}(sqrt3)^{3(frac{x}{y}+frac{y}{x})}=(sqrt3)^{10} \log_2(y-x)(y+x)=log_22^3end{array}right; ; left{begin{array}{ll}3cdot frac{x^2+y^2}{xy}=10\y^2-x^2=8end{array}right; ; left{begin{array}{ll}3x^2+3y^2=10xy\y^2=8+x^2end{array}right$

$y^2=8+x^2; ; to ; ; y=pm sqrt{8+x^2}\\3x^2+3(8+x^2)=10xy; ; ,; ; 6x^2+24=pm 10xcdot sqrt{8+x^2}; ,\\3x^2+12=pm 5xcdot sqrt{8+x^2}; ; ,; ; 9x^4+72x^2+144=25x^2(8+x^2); ,\\16x^4+128x^2-144=0; ,\\x^4+8x^2-9=0; ; to ; ; x^2=1; ,; x^2=-9; (ne; podxodit,; t.k.; x^2geq 0)\\x=pm 1; ; to ; ; y^2=8+1=9; ,; ; y=pm 3\\x_1=1; ,; y_1=3; ,; ; y_1>x_1; ; i; ; y_1>-x_1; ; podxodit\\x_2=1; ,; y_2=-3; ,; ; y_2<x_2; ; ne; podxodit\\x_3=-1; ,; ; y_3=3; ; ,; ; y_3>x_3; ,; y_3>-x_3; ; podxodit$

$x_4=-1; ,; ; y_4=-3; ; ,; ; y_4<x_4; ; ne; podxodit \\Proverka:\\x_1=1; ,; y_1=3:; ; left { {{(sqrt3)^{3cdot frac{10}{3}}=(sqrt3)^{10}=3^5=243} atop {log_24+log_22=2+1=3}} right. ; ; ; verno\\x_3=-1; ,; y_3=3:; ; left { {{(sqrt3})^{3cdot frac{-10}{3}}=(sqrt3)^{-10}=3^{-5}=1/243} atop {log_22+log_24=1+2=3}} right. \\Otvet:; (1,3); .$