Question

A6. Подробно, если можно. Спасибо!

Answer 1

$(\sqrt{4x+\sqrt{16+12x^{2}}})^{2} =(x+2)^{2}\\\\4x+\sqrt{16+12x^{2}}=x^{2}+4x+4\\\\\sqrt{16+12x^{2}}=x^{2}+4\\\\(\sqrt{16+12x^{2}})^{2}=(x^{2}+4)^{2}\\\\16+12x^{2}=x^{4}+8x^{2}+16\\\\x^{4}+8x^{2}-12x^{2}=0\\\\x^{4}-4x^{2}=0$

$x^{2}(x^{2}-4)=0\\\\x^{2}(x-2)(x+2)=0\\\\\left[\begin{array}{ccc}x_{1}=0 \\x-2=0\\x+2=0\end{array}\right\\\\\left[\begin{array}{ccc}x_{1}=0 \\x_{2}=2 \\x_{3}=-2 \end{array}\right$

Проверка:

$1)\sqrt{4*0+\sqrt{16+12*0^{2}}}=0+2\\\\\sqrt{0+\sqrt{16}}=2\\\\\sqrt{4}=2\\\\2=2-verno\\\\2)\sqrt{4*(-2)+\sqrt{16+12*(-2)^{2}}}=-2+2\\\\\sqrt{-8+\sqrt{64}}=0\\\\\sqrt{-8+8}=0\\\\0=0-verno\\\\3)\sqrt{4*2+\sqrt{16+12*2^{2}}}=2+2\\\\\sqrt{8+\sqrt{64}}=4\\\\\sqrt{8+8}=4\\\\\sqrt{16}=4\\\\4=4-verno\\\\Otvet:\boxed{0;-2;2}$

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