ta có \(A=\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

            \(=\sqrt{\frac{1}{x}-\frac{1}{x^2}}+\sqrt{\frac{1}{y}-\frac{2}{y^2}}+\sqrt{\frac{1}{z}-\frac{3}{x^2}}=\sqrt{\frac{1}{4}-\left(\frac{1}{x^2}-2.\frac{1}{2}x+\frac{1}{4}\right)}+\sqrt{\frac{1}{8}-\left(\left(\sqrt{2}y\right)^2-2.\frac{\sqrt{2}}{2\sqrt{2}}x+\frac{1}{8}\right)}+\sqrt{\frac{1}{2}-\left(\left(\sqrt{3}z\right)^2-\frac{1}{z}+\frac{1}{12}\right)}\)

             \(=\sqrt{\frac{1}{4}-\left(\frac{1}{x}-\frac{1}{2}\right)^2}+\sqrt{\frac{1}{8}-\left(\frac{\sqrt{2}}{y}-\frac{1}{2\sqrt{2}}\right)^2}+\sqrt{\frac{1}{12}-\left(\frac{\sqrt{3}}{z}-\frac{1}{2\sqrt{3}}\right)^2}\)

ta có \(\sqrt{\frac{1}{4}-\left(\frac{1}{x}-\frac{1}{2}\right)^2}\le\frac{1}{2}\) ; \(\sqrt{\frac{1}{8}-\left(\frac{\sqrt{2}}{y}-\frac{1}{2\sqrt{2}}\right)^2}\le\frac{1}{2\sqrt{2}}\); \(\sqrt{\frac{1}{12}-\left(\frac{\sqrt{3}}{z}-\frac{1}{2\sqrt{3}}\right)^2}\le\frac{1}{2\sqrt{3}}\)

vậy giá trị lớn nhất của A =\(\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\) khi x=; y=4;z=6

\(P=\dfrac{2x^2-2x+5}{x^2-4x+4}=\dfrac{x^2-4x+4+x^2+2x+1}{x^2-4x+4}=1+\dfrac{\left(x+1\right)^2}{\left(x-2\right)^2}\)Do : \(\dfrac{\left(x+1\right)^2}{\left(x-2\right)^2}\) ≥ 0 ∀x

⇒ \(\dfrac{\left(x+1\right)^2}{\left(x-2\right)^2}\) + 1 ≥ 1

⇒ \(P_{Min}=1\) ⇔ x = - 1

P/s : Day la tim GTNN nha

GTLL hay GTLN Vậy bn

\(A=x^2-4x+3+11\)

\(A=x^2-4x+14\)

\(A=x^2-4x+4+10\)

\(A=\left(x-2\right)^2+10\ge10\)

Dấu = xảy ra khi \(x-2=0\Leftrightarrow x=2\)

vậy \(A_{min}=10\) khi x =2

\(a)A=2+|x+3|\)

Vì \(|x+3|\ge0\)\(\forall x\)

\(\Rightarrow2+|x+3|\ge2\)\(\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy \(Max_A=2\Leftrightarrow x=-3\)

\(b)B=\frac{3}{2}+|2x-1|\)

Vì \(|2x-1|\ge0\)\(\forall x\)

\(\Rightarrow\frac{3}{2}+|2x-1|\ge\frac{3}{2}\)\(\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy \(Max_B=\frac{3}{2}\Leftrightarrow x=\frac{1}{2}\)

Ta có BĐT \(x^2+y^2\ge2xy\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\)

Đẳng thức xảy ra khi \(x-y=0\Leftrightarrow x=y\)

Suy ra \(x+y\ge2\sqrt{xy}\Leftrightarrow2\ge2\sqrt{xy}\Leftrightarrow1\ge xy\)

Đẳng thức xảy ra khi \(x=y=1\)

Vậy GTLN của đơn thức \(xy=1\) khi \(x=y=1\)