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

\(=>P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)

ta có \(\frac{\left(t-2\right)^2}{2}\ge0\left(\forall t\right)\)

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

minP=-1/2

dấu = xảy ra khi x=7/2

a) \(t=\sqrt{2x-3}\ge0\)

<=> \(t^2=2x-3\)

<=> \(x=\frac{t^2+3}{2}\)

=> \(P=\frac{t^2+3}{2}-2t\)

b) khi đó: \(P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}\ge-\frac{1}{2}\)

Dấu "=" xảy ra <=> t = 2  khi đó: x = 7/2

a)đặt t=\(\sqrt{2x-3}\)

=>P=x-2t

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

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

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

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

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

b) \(P=Q\Leftrightarrow2\sqrt{x}+1=x-1\)

\(\Leftrightarrow x-2\sqrt{x}-2=0\)

\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)

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

Vì \(\sqrt{x}-1\ge-1\) \(\Rightarrow\sqrt{x}-1=\sqrt{3}\)

\(\Rightarrow x=\left(\sqrt{3}+1\right)^2=4+2\sqrt{3}\)

Vậy...