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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 $Q_{\min}=4$
\(\left(a^3+b^2+c\right)\left(\dfrac{1}{a}+1+c\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow\dfrac{a^3+b^2+c}{a}\ge\dfrac{\left(a+b+c\right)^2}{1+a+ac}=\dfrac{9}{1+a+ac}\)
\(\Rightarrow\dfrac{a}{a^3+b^2+c}\le\dfrac{1+a+ac}{9}\)
Tương tự: \(\dfrac{b}{b^3+c^2+a}\le\dfrac{1+b+ab}{9}\); \(\dfrac{c}{c^3+a^2+b}\le\dfrac{1+c+bc}{9}\)
Cộng vế:
\(P\le\dfrac{3+a+b+c+ab+bc+ca}{9}\le\dfrac{6+\dfrac{1}{3}\left(a+b+c\right)^3}{9}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(f\left(x\right)\ge0\) ;\(\forall x\in R\)
\(\Leftrightarrow\Delta'=4b^2-ac\le0\)
\(\Leftrightarrow ac\ge4b^2\Rightarrow\sqrt{ac}\ge2b\)
\(F=\dfrac{a+c}{b}\ge\dfrac{2\sqrt{ac}}{b}\ge\dfrac{2.2b}{b}=4\)
\(F_{min}=4\) khi \(a=c=2b\)
b, Ta có : \(0\le x\le1\)
\(\Rightarrow-2\le x-2\le-1< 0\)
Ta có : \(y=f\left(x\right)=2\left(m-1\right)x+\dfrac{m\left(x-2\right)}{\left(2-x\right)}\)
\(=2\left(m-1\right)x-m< 0\)
TH1 : \(m=1\) \(\Leftrightarrow m>0\)
TH2 : \(m\ne1\) \(\Leftrightarrow x< \dfrac{m}{2\left(m-1\right)}\)
Mà \(0\le x\le1\)
\(\Rightarrow\dfrac{m}{2\left(m-1\right)}>1\)
\(\Leftrightarrow\dfrac{m-2\left(m-1\right)}{2\left(m-1\right)}>0\)
\(\Leftrightarrow\dfrac{2-m}{m-1}>0\)
\(\Leftrightarrow1< m< 2\)
Kết hợp TH1 => m > 0
Vậy ...
\(x^2-2\left(m-1\right)x-m^3+\left(m+1\right)^2=0\)
Để pt có hai nghiệm thỏa mãn
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\x_1+x_2=2\left(m-1\right)\le4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m-2\right)\left(m+2\right)\ge0\\m\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m\in\left[-2;0\right]\cup\left(2;+\infty\right)\cup\left\{2\right\}\\m\le3\end{matrix}\right.\)\(\Rightarrow m\in\left[-2;0\right]\cup\left[2;3\right]\)
\(P=x^3_1+x_2^3+x_1x_2\left(3x_1+3x_2+8\right)\)
\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_1\left(x_1+x_2\right)+8x_1x_2\)
\(=8\left(m-1\right)^3+8\left(-m^3+m^2+2m+1\right)\)
\(=-16m^2+40m\)
Vẽ BBT với \(f\left(m\right)=-16m^2+40m\) ;\(m\in\left[-2;0\right]\cup\left[2;3\right]\)
Tìm được \(f\left(m\right)_{min}=-144\Leftrightarrow m=-2\)
\(f\left(m\right)_{max}=16\Leftrightarrow m=2\)
\(\Rightarrow P_{max}=16;P_{min}=-144\)
Vậy....
Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)
Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)
\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)
\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
\(C^1_n+C^2_n=15\)
=>\(n+\dfrac{n!}{\left(n-2\right)!\cdot2!}=15\)
=>\(n+\dfrac{n^2-n}{2}=15\)
=>2n+n^2-n=30
=>n^2+n-30=0
=>n=5
=>(x+2/x^4)^5
SHTQ là: \(C^k_5\cdot x^{5-k}\cdot\left(\dfrac{2}{x^4}\right)^k=C^k_5\cdot x^{5-5k}\cdot2^k\)
SỐ hạng ko chứa x tương ứng với 5-5k=0
=>k=1
=>Số hạng đó là 5*2=10
\(\left\{{}\begin{matrix}x+a+b+c=7\\x^2+a^2+b^2+c^2=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=7-x\\a^2+b^2+c^2=13-x^2\end{matrix}\right.\)
Mà ta có:
\(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow13-x^2\ge\dfrac{\left(7-x\right)^2}{3}\)
\(\Leftrightarrow2x^2-7x+5\le0\)
\(\Leftrightarrow1\le x\le\dfrac{5}{2}\)
Vậy min là 1 khi \(\left\{{}\begin{matrix}x=1\\a=b=c=2\end{matrix}\right.\)
Max là \(\dfrac{5}{2}\) khi \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\a=b=c=\dfrac{3}{2}\end{matrix}\right.\)