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ĐK $-1\leqx\leq \frac{5}{4}$+ Min$P^{2}=6x-3+2\sqrt{(x+1)(5-4x)}\geq 6-3x\geq \frac{9}{4}\Rightarrow P\geq \frac{3}{2}$dấu "=" $\Leftrightarrow x=\frac{4}{5}$+ max$P^{2}=(\frac{1}{2}\sqrt{4x+4}+\sqrt{5-4x})^{2}\leq (\frac{1}{4}+1)(4x+4+5-4x) =\frac{5}{4}.9=\frac{45}{4}$$\Rightarrow P\leq \frac{3\sqrt{5}}{2}$Dấu "=" $\Leftrightarrow x=\frac{-11}{20}$

ĐK $-1 \leq x\leq \frac{5}{4}$+ Min$P^{2}=6x-3+2\sqrt{(x+1)(5-4x)}\geq 6-3x\geq \frac{9}{4}\Rightarrow P\geq \frac{3}{2}$dấu "=" $\Leftrightarrow x=\frac{4}{5}$+ max$P^{2}=(\frac{1}{2}\sqrt{4x+4}+\sqrt{5-4x})^{2}\leq (\frac{1}{4}+1)(4x+4+5-4x) =\frac{5}{4}.9=\frac{45}{4}$$\Rightarrow P\leq \frac{3\sqrt{5}}{2}$Dấu "=" $\Leftrightarrow x=\frac{-11}{20}$

+ ta thấy $P\geq0$ dấu "=" $\Leftrightarrow x=1$ or $x=\frac{4}{5}$+ max$P^{2}=(\frac{1}{2}\sqrt{4x+4}+\sqrt{5-4x})^{2}\leq (\frac{1}{4}+1)(4x+4+5-4x) =\frac{5}{4}.9=\frac{45}{4}$$\Rightarrow P\leq \frac{3\sqrt{5}}{2}$Dấu "=" $\Leftrightarrow x=\frac{-11}{20}$

ĐK $-1\leqx\leq \frac{5}{4}$+ Min$P^{2}=6x-3+2\sqrt{(x+1)(5-4x)}\geq 6-3x\geq \frac{9}{4}\Rightarrow P\geq \frac{3}{2}$dấu "=" $\Leftrightarrow x=\frac{4}{5}$+ max$P^{2}=(\frac{1}{2}\sqrt{4x+4}+\sqrt{5-4x})^{2}\leq (\frac{1}{4}+1)(4x+4+5-4x) =\frac{5}{4}.9=\frac{45}{4}$$\Rightarrow P\leq \frac{3\sqrt{5}}{2}$Dấu "=" $\Leftrightarrow x=\frac{-11}{20}$

+ ta thấy $P\geq0$ dấu "=" $\Leftrightarrow x=1$ or $x=\frac{4}{5}$+ max$P^{2}=(\frac{1}{2}\sqrt{4x+4}+\sqrt{5-4x})^{2}\leq \frac{5}{4}.9=\frac{45}{4}$$\Rightarrow P\leq \frac{3\sqrt{5}}{2}$Dấu "=" $\Leftrightarrow x=\frac{-11}{20}$

+ ta thấy $P\geq0$ dấu "=" $\Leftrightarrow x=1$ or $x=\frac{4}{5}$+ max$P^{2}=(\frac{1}{2}\sqrt{4x+4}+\sqrt{5-4x})^{2}\leq (\frac{1}{4}+1)(4x+4+5-4x) =\frac{5}{4}.9=\frac{45}{4}$$\Rightarrow P\leq \frac{3\sqrt{5}}{2}$Dấu "=" $\Leftrightarrow x=\frac{-11}{20}$