Question

[Trần Xuân Tùng]

Tính:

\(\sqrt{x^4+3x^2}+\sqrt{x^4+6x^2}\)

Answer 1

[Đức Hiếu]

\(\sqrt{x^4+3x^2}+\sqrt{x^4+6x^2}\)

\(=\sqrt{x^4+\dfrac{3}{2}x^2+\dfrac{3}{2}x^2+\dfrac{9}{4}-\dfrac{9}{4}}+\sqrt{x^4+3x^2+3x^2+9-9}\)

\(=\sqrt{\left(x^2+\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2}+\sqrt{\left(x^2+3\right)^2-3^2}\)

\(=\sqrt{\left(x^2+\dfrac{3}{2}-\dfrac{3}{2}\right)\left(x^2+\dfrac{3}{2}+\dfrac{3}{2}\right)}+\sqrt{\left(x^2+3-3\right)\left(x^2+3+3\right)}\)

\(=\sqrt{x^2}.\sqrt{x^2+3}+\sqrt{x^2}.\sqrt{x^2+6}\)

\(=x\left(\sqrt{x^2+3}+\sqrt{x^2+6}\right)\)

Answer 2

[Mashiro Shiina]

\(\sqrt{x^4+3x^2}+\sqrt{x^4+6x^2}\)

\(=\sqrt{x^4+3x^2+\dfrac{9}{4}-\dfrac{9}{4}}+\sqrt{x^4+6x^2+9-9}\)

\(=\sqrt{\left(x^2+\dfrac{3}{2}\right)^2-\dfrac{9}{4}}+\sqrt{\left(x^2+3\right)^2-9}\)

\(=\left|x^2+\dfrac{3}{2}\right|-\dfrac{3}{2}+\left|x^2+3\right|-3\)

Vì: \(\left\{{}\begin{matrix}x^2+\dfrac{3}{2}>0\\x^2+3>0\end{matrix}\right.\)

Nên: \(pt\Leftrightarrow x^2+\dfrac{3}{2}-\dfrac{3}{2}+x^2+3-3\)

\(=2x^2\)

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