1.
\(P=\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\)
\(P^2=\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}+2a^2+2b^2+2c^2\)
Áp dụng BĐT Cô-si :
\(P^2=\frac{1}{2}\cdot\left(\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+\frac{b^2c^2}{a^2}+\frac{a^2b^2}{c^2}+\frac{a^2b^2}{c^2}+\frac{c^2a^2}{b^2}\right)+2\left(a^2+b^2+c^2\right)\)
\(\ge\frac{1}{2}\cdot2\cdot\left(a^2+b^2+c^2\right)+2\cdot\left(a^2+b^2+c^2\right)\)
\(=3\cdot\left(a^2+b^2+c^2\right)=3\)
Do đó \(P\ge\sqrt{3}\)
Dấu đẳng thức xảy ra \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
2. \(x^2+5x+9=\left(x+5\right)\sqrt{x^2+9}\)
Đặt \(\sqrt{x^2+9}=a>0\)
\(\Leftrightarrow a^2=x^2+9\)
\(pt\Leftrightarrow a^2+5x=\left(x+5\right)\cdot a\)
\(\Leftrightarrow a^2+5x-ax-5a=0\)
\(\Leftrightarrow a\left(a-x\right)-5\left(a-x\right)=0\)
\(\Leftrightarrow\left(a-x\right)\left(a-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=x\\a=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+9}=x\\\sqrt{x^2+9}=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+9=x^2\\x^2+9=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\varnothing\\x\in\left\{\pm4\right\}\end{matrix}\right.\)
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