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\(P=\dfrac{1}{\left(x+1\right)^2+5}\le\dfrac{1}{5}\)
\(P_{max}=\dfrac{1}{5}\) 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
Lời giải:
$G=\frac{x^2+x+2}{2x^2-2x+3}$
$\Rightarrow G(2x^2-2x+3)=x^2+x+2$
$\Leftrightarrow x^2(2G-1)-x(2G+1)+(3G-2)=0(*)$
Vì $G$ tồn tại nên dấu "=" tồn tại, điều này có nghĩa là $(*)$ luôn có nghiệm.
$\Rightarrow \Delta=(2G+1)^2-4(2G-1)(3G-2)\geq 0$
$\Leftrightarrow -20G^2+32G-7\geq 0$
$\Leftrightarrow 20G^2-32G+7\leq 0$
$\Leftrightarrow \frac{16+\sqrt{116}}{20}\geq G\geq \frac{16-\sqrt{116}}{20}$
Vậy....
\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)
\(=3+\dfrac{2}{x^2-2x+5}\)
Mà \(x^2-2x+5\ge4\)
=> \(\dfrac{2}{x^2-2x+5}\le\dfrac{1}{2}\)
=> A ≤ 7/2
Dấu "=" xảy ra ⇔ x = 1
Ta có : \(A=\dfrac{3x^2-6x+17}{x^2-2x+5}=\dfrac{3x^2-6x+15+2}{x^2-2x+5}=\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)
\(=3+\dfrac{2}{x^2-2x+5}\)
- Thấy : \(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
Lại có : \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow\dfrac{2}{x^2-2x+5}\le\dfrac{2}{4}=\dfrac{1}{2}\)
\(\Rightarrow3+\dfrac{2}{x^2-2x+5}\le\dfrac{7}{2}\)
\(HayA\le\dfrac{7}{2}\)
Vậy MaxA = \(\dfrac{7}{2}\) Dấu " = " xảy ra <=> x - 1 = 0
<=> x = 1 .
1.
\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)
\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)
\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max
2.
\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)
\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)
\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)
\(E_{min}=-1\) khi \(x=0\)
\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)
\(G_{min}=-2\) khi \(x=2\)
a)
\(A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(A-2=-\dfrac{3}{x^2-8x+22}=-\dfrac{3}{\left(x-4\right)^2+6}\ge-\dfrac{3}{6}=-\dfrac{1}{2}\)
\(A\ge\dfrac{3}{2}\) khi x =4
a/ \(M=\dfrac{x^2-x+1}{x^2+2x+1}=\dfrac{1}{4}+\dfrac{3x^2-6x+3}{x^2+2x+1}=\dfrac{1}{4}+\dfrac{3\left(x-1\right)^2}{x^2+2x+1}\ge\dfrac{1}{4}\)
b/ \(N=\dfrac{3x^2+4x}{x^2+1}=4-\dfrac{x^2-4x+4}{x^2+1}=4-\dfrac{\left(x-2\right)^2}{x^2+1}\le4\)
a. Ta có:\(P\left(x\right)=\dfrac{2x^2-2x+3}{x^2-x+2}=\dfrac{2x^2-2x+4-1}{x^2-x+2}=2-\dfrac{1}{x^2-x+2}\)
Để \(P\left(x\right)\) đạt GTLN thì \(\dfrac{1}{x^2-x+2}\)đạt GTNN
\(\Rightarrow x^2-x+2\) đạt GTNN.
Ta có: \(x^2-x+2=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(\Rightarrow P\left(x\right)=2-\dfrac{1}{x^2-x+2}\ge\dfrac{10}{7}\)
Dấu '' = '' xảy ra khi: \(x=\dfrac{1}{2}\)
Vậy: GTNN của \(P\left(x\right)=\dfrac{10}{7}\) tại \(x=\dfrac{1}{2}\).
\(\dfrac{2\left(x^2-x+2\right)-1}{x^2-x+2}=2-\dfrac{1}{x^2-x+2}\)
ta có \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\) (vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) )
Do đó \(\dfrac{1}{x^2-x+2}\ge\dfrac{1}{\dfrac{7}{4}}=\dfrac{4}{7}\)
Nên P\(\ge2-\dfrac{4}{7}=\dfrac{10}{7}\)
Vậy Min P(x)=\(\dfrac{10}{7}\)