Cái đầu tiên là \(\sqrt[n]{\frac{a_1^n+a_2^n+a_3^n+...+a_n^n}{n}}\)nhé.

ĐKXĐ: \(x\ne0\)

\(\dfrac{9}{x^2}+2+\dfrac{2x}{\sqrt{2x^2+9}}-3=0\)

\(\Leftrightarrow\dfrac{2x^2+9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}-3=0\)

Đặt \(\dfrac{x}{\sqrt{2x^2+9}}=t\)

\(\Rightarrow\dfrac{1}{t^2}+2t-3=0\)

\(\Rightarrow2t^3-3t^2+1=0\)

\(\Rightarrow\left(t-1\right)^2\left(2t+1\right)=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{2x^2+9}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{2x^2+9}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vn\right)\\4x^2=2x^2+9\end{matrix}\right.\)

\(\Rightarrow x=-\dfrac{3\sqrt{2}}{2}\)

`Answer:`

a) \(VT=\)\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2\)

\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

\(=\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\)

\(=a^2.\left(c^2+d^2\right)+b^2.\left(c^2+d^2\right)\)

\(=\left(a^2+b^2\right).\left(c^2+d^2\right)\)

\(=VP\)

b) \(\left(ac+bd\right)^2\le\left(a^2+b^2\right).\left(c^2+d^2\right)\)

\(\Leftrightarrow\left(ac\right)^2+\left(bd\right)^2+2acbd\le\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\)

\(\Leftrightarrow2acbd\le\left(ad\right)^2+\left(bc\right)^2\)

\(\Leftrightarrow\left(ad\right)^2+\left(bc\right)^2-2abdc\ge0\)

\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (Luôn đúng)