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\(A=\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3x^2+6x+9}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)
\(=3+\frac{1}{x^2+2x+3}=3+\frac{1}{\left(x+1\right)^2+2}\le3+\frac{1}{2}=\frac{7}{2}\)
Dấu "=" xảy ra <=> x=-1
Vậy GTLN của A=7/2 khi x=-1
Ta có:
\(A=\frac{3x^2+6x+1}{x^2+2x+3}\)
\(=\frac{3x^2+6x+9}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)
\(=\frac{3\left(x^2+2x+3\right)}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)
\(=3+\frac{1}{x^2+2x+3}\)
Lại có: \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)
\(\Rightarrow\frac{1}{x^2+2x+3}\le\frac{1}{2}\)
\(\Rightarrow A\le3+\frac{1}{2}=\frac{7}{2}\)
Dấu = xảy ra khi \(x^2+2x+3=2\Rightarrow x=-1\)
Vậy \(A_{Min}=\frac{7}{2}\Leftrightarrow x=-1\)
\(A=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ A_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
\(A=x^2-6x+10\)
\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\) \(\forall x\in z\)
\(\Leftrightarrow A_{min}=1khix=3\)
\(B=3x^2-12x+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\) \(\forall x\in z\)
\(\Leftrightarrow B_{min}=-11khix=2\)
\(A=\frac{2x^2+6x+10}{x^2+3x+3}=\frac{2\left(x^2+3x+3\right)+4}{x^2+3x+3}=2+\frac{4}{x^2+3x+3}\)
Để A đạt GTLN thì x2+3x+3 bé nhất
mà x2+3x+3=\(x^2+3.\frac{2}{3}x+\frac{2^2}{3^2}+\frac{23}{9}=\left(x+\frac{2}{3}\right)^2+\frac{23}{9}\ge\frac{23}{9}\)
Dấu "=" xảy ra khi \(x+\frac{2}{3}=0=>x=\frac{-2}{3}\)
lúc đó \(A=2+\frac{4}{\frac{23}{9}}=2+4.\frac{9}{23}=2+\frac{36}{23}=\frac{82}{23}\)
Vậy GTLN của \(A=\frac{82}{23}\)khi \(x=\frac{-2}{3}\)
\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)
\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)
\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)
\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)
\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)
\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)
\(\dfrac{3x^2 + 6x+10}{x^2 + 2x+3}\) \((1) \)
= \(\dfrac{3(x^2+2x+3)+1}{x^2+2x+3}\)
\(= 3+ \dfrac{1}{(x+1)^2 +2}\)
Ta có: \((x+1)^2 \) \(\ge\) \(0\)
\(<=> (x+1)^2 +2\)\(\ge\) \(2\)
\(<=> \dfrac{1}{(x+1)^2 +2}\) \(\le\) \(\dfrac{1}{2}\)
\(<=> 3 + \dfrac{1}{(x+1)^2 +2}\) \(\le\) \(\dfrac{7}{2}\)
Vậy (1) max = \(\dfrac{7}{2}\) \(<=> x = -1 \)
\(\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\frac{1}{x^2+2x+3}=3+\frac{1}{\left(x+1\right)^2+2}\)
Ta có : \(\left(x+1\right)^2+2\ge2\forall x\)
\(\Rightarrow\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
\(\Rightarrow3+\frac{1}{\left(x+1\right)^2+2}\le\frac{7}{2}\forall x\) có GTLN là \(\frac{7}{2}\) tại x = - 1
Vậy .................
\(\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3x^2+6x+6}{x^2+2x+3}+\frac{4}{x^2+2x+3}=\frac{3\left(x^2+2x+3\right)}{x^2+2x+3}+\frac{4}{\left(x^2+2x+1\right)+2}\)
\(=3+\frac{4}{\left(x+1\right)^2+2}\)
Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+2\ge2\Rightarrow\frac{4}{\left(x+1\right)^2+2}\le2\Rightarrow3+\frac{4}{\left(x+1\right)^2+2}\le5\)
=>giá trị nhỏ nhất của biểu thức là 5 <=>(x+1)2=0 <=. x+1=0 <=> x=-1