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\(P=\dfrac{1}{6-4a}+\dfrac{4}{4a}\ge\dfrac{\left(1+2\right)^2}{6-4a+4a}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(P_{min}=\dfrac{3}{2}\) khi \(\dfrac{6-4a}{1}=\dfrac{4a}{2}\Rightarrow a=1\)
Lời giải: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 $Q_{\min}=4$
\(P=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{2a}{b}-\dfrac{2b}{a}-1\)
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
Áp dụng BĐT Cô-si:
\(\dfrac{a^2+1}{a}+\dfrac{a}{a^2+1}\ge2\sqrt{\dfrac{\left(a^2+1\right).a}{a.\left(a^2+1\right)}}=2\)
Vậy Pmin=2\(\Leftrightarrow\dfrac{a^2+1}{a}=\dfrac{a}{a^2+1}\)
\(\Rightarrow a^4+2a^2+1-a^2=0\)
\(\Leftrightarrow\left(a^2+1-a\right)\left(a^2+1+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a^2-a+1=0\\a^2+a+1=0\end{matrix}\right.\)(vô nghiệm)
Vậy Pmin=2.
E = a + \(\dfrac{1}{4a}+b+\dfrac{1}{4b}+c+\dfrac{1}{4c}+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
áp dụng bdt cosi cho cac so duong co:
\(a+\dfrac{1}{4a}\ge2\sqrt{a.\dfrac{1}{4a}}\Leftrightarrow a+\dfrac{1}{4a}\ge1\)
\(b+\dfrac{1}{4b}\ge1,c+\dfrac{1}{4c}\ge1\)
dấu = xảy ra khi a=b=c = 1/2
CM: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{\dfrac{3}{2}}\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6\)\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\)\(\Rightarrow E\ge3+\dfrac{9}{2}\Rightarrow E\ge\dfrac{15}{2}\)
Vậy min E= 15/2 khi a=b=c=1/2
\(P=\left(a^2+4a+12\right)+\left(\dfrac{36a+81}{a^2}+3\right)\)
\(=\left(a+1\right)\left(a+3\right)+\dfrac{3\left(a+9\right)\left(a+3\right)}{a^2}+9\)
\(=\left(a+3\right)\left(\left(a+1\right)+\dfrac{3\left(a+9\right)}{a^2}\right)+9\)
\(=\left(a+3\right)^2\left(a^2-2a+9\right)+9\ge9\)
\("="\Leftrightarrow a=-3\)
cám ơn bn n lắm