اكتب في صيغة (a-b)² أو (a+b)² الأعداد التالية : 11-6√2 ; 12+2√35 ; 5-2√6 ; 14+4√10 ; 27-10√2 ; 27+10√2 اثبت أن: √27+10√2 + √27−10√2 = 10

We need to express the terms under the square roots as perfect squares.
Notice that $27+10\sqrt{2}$ can be written as $(a+b)^2 = a^2+2ab+b^2$.
We have $2ab = 10\sqrt{2}$, so $ab = 5\sqrt{2}$.
If we choose $a=5$ and $b=\sqrt{2}$, then $a^2+b^2 = 5^2 + (\sqrt{2})^2 = 25+2 = 27$.
So, $27+10\sqrt{2} = (5+\sqrt{2})^2$.
Similarly, $27-10\sqrt{2} = (5-\sqrt{2})^2$.

Thus, $\boxed{\sqrt{27+10\sqrt{2}} + \sqrt{27-10\sqrt{2}} = 10}$ is proven.

We need to express the terms under the square roots as perfect squares.
Notice that $14+4\sqrt{10}$ can be written as $(a+b)^2 = a^2+2ab+b^2$.
We have $2ab = 4\sqrt{10}$, so $ab = 2\sqrt{10}$.
If we choose $a=2$ and $b=\sqrt{10}$, then $a^2+b^2 = 2^2 + (\sqrt{10})^2 = 4+10 = 14$.
So, $14+4\sqrt{10} = (2+\sqrt{10})^2$.
Similarly, $14-4\sqrt{10} = (2-\sqrt{10})^2$.

Since $2-\sqrt{10}$ is negative (because $\sqrt{10} \approx 3.16$), $\sqrt{(2-\sqrt{10})^2} = |2-\sqrt{10}| = -(2-\sqrt{10}) = \sqrt{10}-2$.

Thus, $\boxed{\sqrt{14+4\sqrt{10}} + \sqrt{14-4\sqrt{10}} = 2\sqrt{10}}$ is proven.