4x+5y + 20xy=5 16x2 +25y2 + 20xy=7

Question 1

4x+5y + 20xy=5
16x2 +25y2 + 20xy=7

Question 2

Да

Question 3

$\left \{ {{4x+5y+20xy=5} \atop {16x^2+25y^2+20xy=7}} \right. \\\\a)\; \; 16x^2+25y^2+20xy=7\\\\\star \; (4x)^2+(5y)^2+2\cdot 4x\cdot 5y=(4x)^2+(5y)^2+40xy=(4x+5y)^2\; \; \Rightarrow \\\\16x^2+25y^2+20xy=(4x+5y)^2-20xy\; \; \Rightarrow \\\\(4x+5y)^2-20xy=7\; \; \Rightarrow \; \; (4x+5y)^2=7+20xy\\\\b)\; \; 4x+5y+20xy=5\; \; \Rightarrow \; \; 20xy=5-4x-5y\\\\c)\; \; (4x+5y)^2=7+(5-4x-5y)\\\\(4x+5y)^2=12-(4x+5y)\\\\t=4x+5y\; \; \; \Rightarrow \; \; t^2+t-12=0\; ,\\\\t_1=-4\; ,\; t_2=3\; \;(teorema\; Vieta)$

$d)\; \; \left \{ {{4x+5y+20xy=5} \atop {4x+5y=-4}} \right. \; \left \{ {{-4+20xy=5} \atop {4x+5y=-4}} \right. \; \left \{ {{20xy=9} \atop {4x+5y=-4}} \right. \\\\ \left \{ {{16x^2+25y^2+20xy=7} \atop {20xy=9}} \right. \; \left \{ {{16x^2+20y^2+9=7} \atop {20xy=9}} \right. \; \left \{ {{16x^2+25y^2=-2} \atop {20xy=9}} \right.$

Система не имеет решений, т.к. 16х²+25у²≥0 при любых значениях переменных х и у, а в 1 уравнении системы сумма квадратов получилась отрицательной.

$e)\; \; \left \{ {{20xy=5-(4x+5y)} \atop {4x+5y=3}} \right. \; \left \{ {{20xy=5-3} \atop {4x+5y=3}} \right. \; \left \{ {{20xy=2} \atop {4x+5y=3}} \right. \\\\ \left \{ {{16x^2+25y^2+20xy=7} \atop {20xy=2}} \right. \; \left \{ {{16x^2+25y^2+2=7} \atop {20xy=2}} \right. \; \left \{ {{16x^2+25y^2=5} \atop {20xy=2}} \right. \\\\ \left \{ {{16x^2+25y^2=5} \atop {5y=3-4x}} \right. \; \left \{ {{16x^2+(3-4x)^2=5} \atop {5y=3-4x}} \right. \; \left \{ {{16x^2+9-24x+16x^2=5} \atop {5y=3-4x}} \right.$

$32x^2-24x+4=0\; |:4\\\\8x^2-6x+1=0\\\\D/4=3^2-8=1\\\\x_1= \frac{3-1}{8}= \frac{1}{4} \; ,\; \; x_2= \frac{3+1}{8}=\frac{1}{2} \\\\y_1= \frac{3-4x}{5}= \frac{3-1}{5}= \frac{2}{5}\; ,\; \; y_2= \frac{3-4x}{5}= \frac{3-2}{5}=\frac{1}{5}\\\\Otvet:\; \; ( \frac{1}{4}\; , \frac{2}{5})\; ,\; \; ( \frac{1}{2}\; ,\; \frac{1}{5})\; .$