\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(f\left(x\right)=\left(2-x\right)\left(x+3\right)\le\dfrac{1}{4}\left(2-x+x+3\right)^2=\dfrac{25}{4}\)

\(f\left(x\right)_{max}=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

\(A=\sqrt{x^2+y\left(y-2x\right)}+\sqrt{y^2+z\left(z-2y\right)}+\sqrt{x^2+z\left(z-2x\right)}\)

\(=\sqrt{x^2-2xy+y^2}+\sqrt{y^2-2yz-z^2}+\sqrt{x^2-2xz+z^2}\)

\(=\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-z\right)^2}+\sqrt{\left(z-x\right)^2}\)

\(=x-y+y-z+z-x\)

\(=0\)

A=(6-2x)(12-3y)(2x+3y)/6

<=(6-2x+12-3y+2x+3y)³/(6.27)

=18³/(6.27)=36

Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Đặt a+b=x;b+c=y;c+a=z

\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)

Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)

a: ĐKXĐ: (x+4)(x-1)<>0

hay \(x\notin\left\{-4;1\right\}\)

b: \(y-3=\dfrac{2x^2+6\sqrt{\left(x^2+1\right)\left(x-2\right)}+5-3x^2-9x+12}{x^2+3x-4}\)

\(=\dfrac{-x^2-9x+17+6\sqrt{\left(x^2+1\right)\left(x-2\right)}}{x^2+3x-4}< =0\)

=>y<=3

a.

\(y=x^2\left(4-2x\right)=x.x.\left(4-2x\right)\le\left(\dfrac{x+x+4-2x}{3}\right)^3=\dfrac{64}{27}\)

\(y_{max}=\dfrac{64}{27}\) khi \(x=4-2x\Rightarrow x=\dfrac{4}{3}\)

b.

\(y=x\left(2-x\right)^2=\dfrac{1}{2}.2x.\left(2-x\right)\left(2-x\right)\le\dfrac{1}{2}\left(\dfrac{2x+2-x+2-x}{3}\right)^3=\dfrac{32}{27}\)

\(y_{max}=\dfrac{32}{27}\) khi \(2x=2-x\Rightarrow x=\dfrac{2}{3}\)