Question

Нужно решить 3 неравенства (4,5,6), точно так, как на примере.

Answer 1

Объяснение:

4)

$\displaystyle\\x^2+9y^2+2x+6y+2\geq 0\\\\x^2+9y^2+2x+6y+2=\\\\=(x^2+2x+1)+(9y^2+6y+1)=\\\\=(x+1)^2+(3y+1)^2\geq 0$

ч.т.д

5)

$x^2-6xy+10y^2-4y+70\\\\x^2-6xy+10y^2-4y+7=\\\\=(x^2-6xy+9y^2)+(y^2-4y+4)+3=\\\\=(x-3y)^2+(y-2)^2+30$

ч.т.д.

6)

$\displaystyle\\\frac{x^2+7}{\sqrt{x^2+6} } \geq 2\ \ \ \ \ \ \Big|\cdot\sqrt{x^2+6}0 \\\\\\x^2+7\geq 2\sqrt{x^2+6}\\\\\\x^2+7-2\sqrt{x^2+6}=x^2+6+1- 2\sqrt{x^2+6}=\\\\\\=x^2+6-2\sqrt{x^2+6}+1=\\\\\\=\Big(\sqrt{x^2+6}\Big)^2-2\sqrt{x^2+6}+1^2=\\\\\\=\Big(\sqrt{x^2+6}-1\Big)^2\geq 0$

ч.т.д.