Lời giải:
\(t=\sqrt{2x-3}\Rightarrow t^2=2x-3\Rightarrow x=\frac{t^2+3}{2}\)

Khi đó:

\(P=x-2\sqrt{2x-3}=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}\)

a)đặt t=\(\sqrt{2x-3}\)

=>P=x-2t

=>t=\(\frac{x-P}{2}\)

a.

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

                    \(\Rightarrow y^2=x+2\)

                    \(\Rightarrow x=y^2-2\)

thay vào A ta có:\(A=x-2\sqrt{x+2}\)

\(\Rightarrow A=y^2-2y=y^2-2y-2\)

b.

\(A=x-2\sqrt{x+2}\)

Điều kiện:x+2≥0⇔x>-2

ta có:\(A=x-2\sqrt{x+2}\)

            \(=\left(x+2\right)-2\sqrt{x+2}.1+1-3\)

            \(=\left(\sqrt{x+12}-1\right)^2-3\)

vì \(\left(\sqrt{x+2}-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(\sqrt{x+2}-1\right)^2-3\ge-3\forall x\)

vậy GTNN của A là-3

a/ y=\(\sqrt{x+2}\)→\(y^2-2=x\)

⇒A=\(y^2-2-2y\)

b/ A=\(y^2-2y-2\)=\(\left(y^2-2y+1\right)-3\)=\(\left(y-1\right)^2-3\)≥ -3

⇒\(A_{min}=-3\)

dấu = xảy ra khi y=1⇒x= -1

a, ĐKXĐ : \(2x-3\ge0\)

=> \(x\ge\frac{3}{2}\)

Ta có : \(P=x-2\sqrt{2x-3}\)

- Đặt \(t=\sqrt{2x-3}\left(t\ge0\right)\)

=> \(t^2=2x-3\)

=> \(x=\frac{t^2+3}{2}\)

- Thay vào P ta được : \(P=\frac{t^2+3}{2}-2t\)

b, Ta có : \(P=\frac{t^2+3-4t}{2}\)

=> \(P=\frac{t^2-4t+4-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)

Ta thấy : \(\left(t-2\right)^2\ge0\forall x\)

=> \(\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\ge-\frac{1}{2}\forall x\)

Vậy \(Min_P=-\frac{1}{2}\) <=> \(t-2=0\)

<=> \(t=2\left(TM\right)\)

<=> \(\sqrt{2x-3}=2\)

<=> \(2x-3=4\)

<=> \(2x=7\)

<=> \(x=\frac{7}{2}\left(TM\right)\)

ĐKXĐ:...

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

\(B=\frac{2x\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2x}{\sqrt{x}}=2\sqrt{x}\)

\(A=x+1=\sqrt{4+\sqrt{7}}+1=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+1=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}+1=\frac{1+\sqrt{14}+\sqrt{2}}{2}\)

\(B< -x+3\Leftrightarrow2\sqrt{x}< -x+3\Leftrightarrow x+2\sqrt{x}-3< 0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)< 0\Leftrightarrow\sqrt{x}-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)

Ta có:

\(A-B=x+1-2\sqrt{x}=\left(\sqrt{x}-1\right)^2\ge0\) \(\forall x\in TXĐ\)

Mà \(x\ne1\Rightarrow\) dấu "=" ko xảy ra

\(\Rightarrow A-B>0\Rightarrow A>B\)