Biết `a-b =` $\sqrt{2}+1$, `b-c=` $\sqrt{2}-1$. Tính $A=a^{2}+b^{2}+c^{2}-ab-bc-ca$

Ta có:

$\left.\begin{matrix} a-b=\sqrt{2}+1\\b-c=\sqrt{2}-1 \end{matrix}\right\}$ $\Rightarrow$ $a-b+b-c=\sqrt{2}+1+\sqrt{2}-1$

$\Leftrightarrow$ $a-c=2\sqrt{2}$

$A=a^{2}+b^{2}+c^{2}-ab-bc-ca$

$\Leftrightarrow$ $2A=2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$

$\Leftrightarrow$ $2A=(a^{2}-2ab+b^{2})+(b^{2}-2bc+c^{2})+(a^{2}-2ac+c^{2})$

$\Leftrightarrow$ $2A=\sqrt{(2+1)^{2}} + \sqrt{(2-1)^{2}} + (2\sqrt{2})^{2}$

$\Leftrightarrow$ $2A=2+2\sqrt{2}+1+2-2\sqrt{2}+1+8$

$\Leftrightarrow$ $2A=14$

$\Longrightarrow$ $A=7$

Vậy `A=7`