\(\Leftrightarrow Ax^2-2A=-7x^2+6x+3\\ \Leftrightarrow x^2\left(A+7\right)-6x-2A-3=0\\ \Leftrightarrow\Delta'=3^2+\left(2A+3\right)\left(A+7\right)\ge0\\ \Leftrightarrow2A^2+17A+30\ge0\\ \Leftrightarrow\left[{}\begin{matrix}A\le-6\\A\ge-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow A\text{ ko có max và min}\)

a) \(A=\frac{3x^2+6x+10}{x^2+2x+3}\)

\(A=\frac{3x^2+6x+9+1}{x^2+2x+3}\)

\(A=\frac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}\)

\(A=\frac{3\left(x^2+2x+3\right)}{x^2+2x+3}+\frac{1}{x^2+2x+1+2}\)

\(A=3+\frac{1}{^{\left(x+1\right)^2+2}}\le3+\frac{1}{2}=\frac{7}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-1\)

sai

\(A=-x^2-6x+1\)

\(=-x^2-6x-9+10\)

\(=-\left(x^2+2\cdot x\cdot3+3^2\right)+10\)

\(=-\left(x+3\right)^2+10\)

Ta có: \(\left(x+3\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x+3\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+3\right)^2+10\le10\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy \(Max_A=10\) khi \(x=-3\)

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) 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 \(x=-6\)

\(x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=3\)

Biểu thức không có GTLN

a:

ĐKXĐ: x<>-10

\(A=\frac{x}{\left(x+10\right)^2}\)

\(=\frac{x}{x^2+10x+25}=\frac{1}{x+10+\frac{25}{x}}\)

\(x+\frac{25}{x}\ge2\cdot\sqrt{x\cdot\frac{25}{x}}=2\cdot5=10\forall x\) thỏa mãn ĐKXĐ

=>\(x+\frac{25}{x}+10\ge10+10=20\forall x\) thỏa mãn ĐKXĐ
=>\(A=\frac{1}{x+\frac{25}{x}+10}\le\frac{1}{20}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(x=\frac{25}{x}\)

=>\(x^2=25\)

=>x=5(nhận) hoặc x=-5(nhận)

b: ĐKXĐ: x<>1/2

Đặt y=2x-1

=>2x=y+1

=>\(x=\frac{y+1}{2}\)

\(4x^2-6x+1=(2x)^2-3\cdot2x+1\)

\(=(y+1)^2-3(y+1)+1=y^2-y-1\)

\(B = \frac{y^2 - y - 1}{y^2} = 1 - \frac{1}{y} - \frac{1}{y^2}\)

\(=-\frac{1}{y^2}-\frac{1}{y}-\frac14+\frac54=-\left(\frac{1}{y}+\frac12\right)^2+\frac54\le\frac54\forall y\)

Dấu '=' xảy ra khi \(\frac{1}{y}+\frac12=0\)

=>y=-2

=>\(x=\frac{y+1}{2}=\frac{-2+1}{2}=-\frac12\)