$\left\{\begin{matrix} x^2+y^2+xy=9 & & \\ y^2+z^2+yz=16 & & \\ z^2+x^2+zx=25 & & \end{matrix}\right.$

Bài toán:

$\left\{\begin{matrix} x^2+y^2+xy=9 & & \\ y^2+z^2+yz=16 & & \\ z^2+x^2+zx=25 & & \end{matrix}\right.$

Tìm $xy+yz+xz$

Lời giải:

$(1)+(2)+(3)\Leftrightarrow 50=2\sum a^2+\sum ab$

$\Leftrightarrow 50-\left ( \sum a \right )^2=\sum a^2-\sum ab~~(*)$

$(1)-(2)\Leftrightarrow (a-c)(a+b+c)=-7$

$(2)-(3)\Leftrightarrow (b-a)(a+b+c)=-9$

$(1)-(3)\Leftrightarrow (b-c)(a+b+c)=-16$

$\Rightarrow \dfrac{1}{a+b+c}=\dfrac{a-b}{9}=\dfrac{c-a}{7}=\dfrac{c-b}{16}~~(**)$

Từ $(*)$ và $(**)$ ta có: $50-\left ( \sum a \right )^2=\dfrac{1}{2}\left [ \sum (a-b)^2 \right ]$

$=\dfrac{1}{2}\left [ \left (\dfrac{9}{a+b+c}  \right )^2+\left ( \dfrac{-16}{a+b+c} \right )^2+\left ( \dfrac{-7}{a+b+c} \right )^2 \right ]$

Đặt $(a+b+c)^2=x\Rightarrow 50-x=\dfrac{1}{2}\left ( \dfrac{81}{x}+\dfrac{256}{x}+\dfrac{49}{x} \right )$

$$\Leftrightarrow 2x^2-100x+386=0$$

$$\Leftrightarrow \begin{bmatrix}x=25+12\sqrt{3}  &  & \\ x=25-12\sqrt{3}  &  &  \end{bmatrix}$$

Có: $$2(a+b+c)^2-50=2(a+b+c)^2-\left [ 2\sum a^2+\sum ab \right ]=3\sum ab$$

$$\Rightarrow \sum ab=\dfrac{2x^2-50}{3}=\pm 8\sqrt{3}$$