Question

$1) x^{2} -5\sqrt{3}x+18=0$$2) (3x+1)^{2}-(3x-1)^{2}=6$$3)\left \{ {{x^{2} -y^{2} =25} \atop {x-2y=5}} \right.$

Answer 1

Ответ:

$2\sqrt{3}, \quad 3\sqrt{3};$

$0,5;$

$(-8\frac{1}{3}; -6\frac{2}{3}), \quad (5; 0);$

Объяснение:

$1) \quad x^{2}-5\sqrt{3}x+18=0;$

Решаем уравнение при помощи теоремы Виета:

$\left \{ {{x_{1}+x_{2}=-(-5\sqrt{3})} \atop {x_{1} \cdot x_{2}=18}} \right. \Leftrightarrow \left \{ {{x_{1}+x_{2}=5\sqrt{3}} \atop {x_{1} \cdot x_{2}=18}} \right. \Leftrightarrow \left \{ {{x_{1}=2\sqrt{3}} \atop {x_{2}=3\sqrt{3}}} \right. ;$

$2) \quad (3x+1)^{2}-(3x-1)^{2}=6;$

$9x^{2}+6x+1-(9x^{2}-6x+1)=6;$

$9x^{2}-9x^{2}+6x+6x+1-1=6;$

$12x=6;$

$x=0,5;$

$3) \quad \left \{ {{x^{2}-y^{2}=25} \atop {x-2y=5}} \right. \Leftrightarrow \left \{ {{(2y+5)^{2}-y^{2}=25} \atop {x=2y+5}} \right. \Leftrightarrow \left \{ {{4y^{2}+20y+25-y^{2}-25=0} \atop {x=2y+5}} \right. \Leftrightarrow \left \{ {{3y^{2}+20y=0} \atop {x=2y+5}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{y(3y+20)=0} \atop {x=2y+5}} \right. \Leftrightarrow \left \{ {{y=0} \atop {x=2 \cdot 0+5}} \right. \vee \left \{ {{3y+20=0} \atop {x=2y+5}} \right. \Leftrightarrow \left \{ {{x=5} \atop {y=0}} \right. \vee \left \{ {{y=-6\frac{2}{3}} \atop {x=2y+5}} \right. \Leftrightarrow \left \{ {{x=5} \atop {y=0}} \right. \vee \left \{ {{x=-8\frac{1}{3}} \atop {y=-6\frac{2}{3}}} \right.$

$(-8\frac{1}{3}; -6\frac{2}{3}), \quad (5; 0);$