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\(x>0\)
\(C=x+\dfrac{1}{4x}+\dfrac{x}{\left(2x+1\right)^2}=\dfrac{4x^2+1}{4x}+\dfrac{x}{\left(2x+1\right)^2}\)
-Ta đặt \(A=T=4x^2+1;B=4x\) thì ta có:
\(A\ge B\Rightarrow A+T\ge B+T\) (do \(T>0\))\(\Rightarrow\dfrac{A+T}{B+T}\ge1\)
-Do đó: \(C=\dfrac{4x^2+1}{4x}+\dfrac{x}{\left(2x+1\right)^2}\ge\text{}\dfrac{4x^2+1+4x^2+1}{4x+4x^2+1}+\dfrac{x}{\left(2x+1\right)^2}=\dfrac{2\left(4x^2+1\right)}{\left(2x+1\right)^2}+\dfrac{8x}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=\dfrac{2\left(2x+1\right)^2}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=2-\dfrac{7x}{\left(2x+1\right)^2}\)
-Áp dụng BĐT AM-GM ta có:
\(C\ge2-\dfrac{7x}{\left(2x+1\right)^2}\ge2-\dfrac{7x}{4.2x}=2-\dfrac{7}{8}=\dfrac{9}{8}\)
\(C=\dfrac{9}{8}\Leftrightarrow x=\dfrac{1}{2}\)
-Vậy \(C_{min}=\dfrac{9}{8}\)
\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
Có : \(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)\(\ge\left|x^2-x+1-x^2+x-2\right|=\left|-1\right|=1\)
Vậy Pmin=1\(\Leftrightarrow\left(x^2-x+1\right)\left(-x^2+x-2\right)\ge0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2-x+2\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x+1\ge0\\x^2-x+2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x+1\le0\\x^2-x+2\ge0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\in R\\x\in\varnothing\end{matrix}\right.\\\left\{{}\begin{matrix}x\in\varnothing\\x\in R\end{matrix}\right.\end{matrix}\right.\)
Vậy không tồn tại GTNN của P.
\(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)
\(P=\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right|+\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{7}{4}\right|\)
\(P=\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right|+\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right|\)
\(P=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{10}{4}\ge\dfrac{10}{4}=\dfrac{5}{2}\)
\(\Rightarrow P_{min}=\dfrac{5}{2}\) khi \(x=\dfrac{1}{2}\)
(a) Điều kiện : \(x\ne-1.\)
Ta có : \(P=\dfrac{x^4+x}{x^2-x+1}+1-\dfrac{2x^2+3x+1}{x+1}\)
\(=\dfrac{x\left(x^3+1\right)}{x^2-x+1}+1-\dfrac{\left(2x+1\right)\left(x+1\right)}{x+1}\)
\(=\dfrac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}+1-\left(2x+1\right)\)
\(=x\left(x+1\right)+1-2x-1\)
\(=x^2-x.\)
Vậy : Với mọi \(x\ne-1\) thì \(P=x^2-x.\)
(b) Ta có : \(P=x^2-x\)
\(=\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy : \(MinP=-\dfrac{1}{4}.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=\dfrac{1}{2}.\)