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

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

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

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

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 \(x=-6\)

A = -(x²+6x-11)

=-(x²+6x+9-20)

=-(x+3)² + 20 \(\le20\)

vậy min A = 20

dấu = xảy ra khi x = -3

câu B bạn xem có nhầm đề hay thiếu gì k thì bổ sung nhé

à tớ nhầm 1 chỗ, là max A = 20

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

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

a) Ta có : \(A=-6x+x^2+11\)

\(\Rightarrow A=\left(x^2-6x+9\right)+2\)

\(\Rightarrow A=\left(x-3\right)^2+2\ge2\)

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

Vậy \(minA=2\Leftrightarrow x=3\)

b) \(B=-1+2x^x+10x\)

\(\Rightarrow\)Tớ đang thắc mắc cái chỗ 2x^x :)))