\(P=\left(a^2+4a+12\right)+\left(\dfrac{36a+81}{a^2}+3\right)\)

\(=\left(a+1\right)\left(a+3\right)+\dfrac{3\left(a+9\right)\left(a+3\right)}{a^2}+9\)

\(=\left(a+3\right)\left(\left(a+1\right)+\dfrac{3\left(a+9\right)}{a^2}\right)+9\)

\(=\left(a+3\right)^2\left(a^2-2a+9\right)+9\ge9\)

\("="\Leftrightarrow a=-3\)

giúp mình với 

\(P=\dfrac{1}{6-4a}+\dfrac{4}{4a}\ge\dfrac{\left(1+2\right)^2}{6-4a+4a}=\dfrac{9}{6}=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(\dfrac{6-4a}{1}=\dfrac{4a}{2}\Rightarrow a=1\)

Lời giải:

$a^2-2ab-3b^2\geq 0$

$\Leftrightarrow (a^2+ab)-(3ab+3b^2)\geq 0$

$\Leftrightarrow a(a+b)-3b(a+b)\geq 0$

$\Leftrightarrow (a+b)(a-3b)\geq 0$

$\Leftrightarrow a-3b\geq 0$ (do $a+b>0$ với mọi $a,b>0$)

$\Leftrightarrow a\geq 3b$

Xét hiệu:

$P-\frac{37}{3}=\frac{4a^2+b^2}{ab}-\frac{37}{3}$

$=\frac{12a^2+3b^2-37ab}{3ab}=\frac{(a-3b)(12a-b)}{3ab}\geq 0$ do $a\geq 3b>0$

$\Rightarrow P\geq \frac{37}{3}$

Vậy $P_{\min}=\frac{37}{3}$

Lời giải:Vì $f(x)\geq 0$ nên $\Delta=b^2-4ac\leq 0$

$\Leftrightarrow 4ac\geq b^2$

Áp dụng BĐT AM-GM:

$Q=\frac{4a+c}{b}\geq \frac{4\sqrt{ac}}{b}\geq \frac{4\sqrt{b^2}}{b}=\frac{4b}{b}=4$

Vậy $Q_{\min}=4$

=>\(25\cdot\dfrac{\sqrt{a-3}}{5}-7\cdot\dfrac{2}{3}\cdot\sqrt{a-3}-7\sqrt{a^2-9}+18\cdot\dfrac{1}{3}\sqrt{a^2-9}=0\)

=>\(\sqrt{a-3}\cdot\dfrac{1}{3}-\sqrt{a^2-9}=0\)

=>\(\sqrt{a-3}\left(\dfrac{1}{3}-\sqrt{a+3}\right)=0\)

=>a-3=0 hoặc a+3=1/9

=>a=3 hoặc a=-26/9