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Lời giải:
\(A=\frac{x^2+x+1}{x^2+2x+1}=\frac{x^2+2x+1-x}{x^2+2x+1}=1-\frac{x}{x^2+2x+1}=1-\frac{x}{(x+1)^2}\)
Ta thấy \((x+1)^2-4x=x^2-2x+1=(x-1)^2\geq 0\)
\(\Rightarrow (x+1)^2\geq 4x\Rightarrow \frac{x}{(x+1)^2}\leq \frac{x}{4x}=\frac{1}{4}\)
\(\Rightarrow A=1-\frac{x}{(x+1)^2}\geq 1-\frac{1}{4}=\frac{3}{4}\)
Vậy \(A_{\min}=\frac{3}{4}\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\), tức là A đạt min khi $x=1$
Ta có: \(Q=\dfrac{x^2+x+1}{x^2+2x+1}\)
\(\Rightarrow\dfrac{1}{Q}=\dfrac{x^2+2x+1}{x^2+x+1}\)
Để Q min thì \(\dfrac{1}{Q}\) max
\(\dfrac{1}{Q}=\dfrac{x^2+2x+1}{x^2+x+1}=1+\dfrac{x}{x^2+x+1}\)
\(=1+\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{-x^2+2x+1}{x^2+x+1}=\dfrac{4}{3}-\dfrac{1}{3}.\dfrac{\left(-x-1\right)^2}{x^2+x+1}\le\dfrac{4}{3}\)
( Vì mẫu > 0 và tử \(\ge0\) )
\(\Rightarrow\dfrac{1}{Q}\) đạt GTNN là \(\dfrac{4}{3}\) khi x =1
Vậy Q đạt GTNN là \(\dfrac{3}{4}\) khi x = 1
Ta có: \(\dfrac{a+b}{a}=\dfrac{a}{b}\)
\(\Leftrightarrow\dfrac{a}{b}-1-\dfrac{1}{\dfrac{a}{b}}=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}\right)^2-\dfrac{a}{b}-1=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{\sqrt{5}+1}{2}\\\dfrac{a}{b}=\dfrac{-\sqrt{5}+1}{2}\end{matrix}\right.\)
Thế \(\dfrac{a}{b}\) vào PT \(x^2-x-1\)
\(\Rightarrowđpcm\)
\(P=\frac{1}{x^2+2x+6}\)
\(P=\frac{1}{\left(x+1\right)^2+5}\ge\frac{1}{5}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy Pmin = 1/5 khi và chỉ khi x = -1
Lời giải:
$x^2+2x+6=(x^2+2x+1)+5=(x+1)^2+5\geq 5$ với mọi $x\in\mathbb{R}$
Do đó: $P=\frac{1}{x^2+2x+6}\leq \frac{1}{5}$
Vậy $P_{\max}=\frac{1}{5}$. Giá trị đạt tại $x=-1$
\(P=\dfrac{1}{\left(x+1\right)^2+5}\le\dfrac{1}{5}\)
\(P_{max}\) khi \(x+1=0\Leftrightarrow x=-1\)
a: \(M=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^2+1}{2}\)
\(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
ĐKXĐ: \(x\ne1\)
\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)
\(\Rightarrow C=\left[2x^2+1-1\left(x^2+x+1\right)\right]\div\left[\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2\right)\right]\)
\(\Rightarrow C=(2x^2+1-x^2-x-1)\div\left[\left(x-1\right)\left(x^2+x+1-x^2+2\right)\right]\)
\(\Rightarrow C=\left(x^2-x\right)\div\left[\left(x-1\right)\left(x+3\right)\right]\)
\(P=\dfrac{x^2+x+1}{x^2+2x+1}\) ( x # -1)
\(P=\dfrac{\left(x+1\right)^2-x}{\left(x+1\right)^2}\)
\(P=1-\dfrac{x}{\left(x+1\right)^2}\)
\(P=1+\dfrac{1}{\left(x+1\right)^2}-\dfrac{1}{x+1}\)
\(P=\left[\dfrac{1}{\left(x+1\right)^2}-2.\dfrac{1}{x+1}.\dfrac{1}{2}+\dfrac{1}{4}\right]+1-\dfrac{1}{4}\)
\(P=\left(\dfrac{1}{x+1}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do : \(\left(\dfrac{1}{x+1}-\dfrac{1}{2}\right)^2\) ≥ 0 ∀x # -1
⇒ \(\left(\dfrac{1}{x+1}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) ≥ \(\dfrac{3}{4}\)
⇒ PMIN = \(\dfrac{3}{4}\) ⇔ x + 1 = 2 ⇔ x = 1
Mk làm cách khác nhé !!!
P = \(\dfrac{x^2+x+1}{x^2+2x+1}\)
P - 1 = \(\dfrac{x^2+x+1}{x^2+2x+1}\) - 1
P - 1 = \(\dfrac{-x}{x^2+2x+1}=\dfrac{-x}{x\left(x+2+\dfrac{1}{x}\right)}\)
P - 1 = \(\dfrac{-1}{x+\dfrac{1}{x}+2}\)
P - 1 = \(\dfrac{-1}{\left(\sqrt{x}-\sqrt{\dfrac{1}{x}}\right)^2+4}\) ≥ \(\dfrac{-1}{4}\)
⇒ P ≥ 1 - \(\dfrac{1}{4}=\dfrac{3}{4}\)
⇒ PMin = \(\dfrac{3}{4}\)
Dấu"=" xảy ra khi và chỉ khi : \(x=\dfrac{1}{x}\) ⇔ x = 1
\(P=\dfrac{1}{\left(x+1\right)^2+5}\le\dfrac{1}{5}\)
\(P_{max}=\dfrac{1}{5}\) khi \(x+1=0\Rightarrow x=-1\)
\(Q=\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{4x^2+4x+4}{4\left(x+1\right)^2}=\dfrac{3\left(x^2+2x+1\right)+x^2-2x+1}{4\left(x+1\right)^2}=\dfrac{3}{4}+\dfrac{\left(x-1\right)^2}{4\left(x+1\right)^2}\)
\(Q_{min}=\dfrac{3}{4}\) khi \(x-1=0\Rightarrow x=1\)
1: \(x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5>=5\forall x\)
=>\(P=\dfrac{1}{x^2+2x+6}< =\dfrac{1}{5}\forall x\)
Dấu '=' xảy ra khi x+1=0
=>x=-1