a/ ĐKXĐ: $x\ge 0$
$P\,=\dfrac{-x}{\sqrt x+1}\\\quad =\dfrac{-x-2\sqrt x-1+2\sqrt x+2-1}{\sqrt x+1}\\\quad =\dfrac{-(x+2\sqrt x+1)+(2\sqrt x+2)-1}{\sqrt x+1}\\\quad =\dfrac{-(\sqrt x+1)^2+2(\sqrt x+1)-1}{\sqrt x+1}\\\quad =-(\sqrt{x}+1)+2-\dfrac{1}{\sqrt x+1}\\\quad =-\left[(\sqrt x+1)+\dfrac{1}{\sqrt x+1}\right]+2$
Áp dụng BĐT Cô si với 2 số không dương $\sqrt x+1,\dfrac{1}{\sqrt x+1}$
$\sqrt x+1+\dfrac{1}{\sqrt x+1}\ge 2\sqrt{(\sqrt x+1).\dfrac{1}{\sqrt x+1}}\\↔\sqrt x+1+\dfrac{1}{\sqrt x+1}\ge 2\\↔-\left[(\sqrt x+1)+\dfrac{1}{\sqrt x+1}\right]\le -2\\↔P\le 0\\→\max P=0$
$→$ Dấu "=" xảy ra khi $\sqrt x+1=\dfrac{1}{\sqrt x+1}$
$↔(\sqrt x+1)^2=1$
mà $\sqrt x+1\ge 1>0$
$→\sqrt x+1=1\\↔\sqrt x=0\\↔x=0(TM)$
Vậy $\max P=0$ khi $x=0$
b/ ĐKXĐ: $x\ge 0$
$P\,=\dfrac{-x-3}{\sqrt x+1}\\\quad =\dfrac{-x-2\sqrt x-1+2\sqrt x+2-4}{\sqrt x+1}\\\quad =\dfrac{-(x+2\sqrt x+1)+(2\sqrt x+2)-4}{\sqrt x+1}\\\quad =\dfrac{-(\sqrt x+1)^2+2(\sqrt x+1)-4}{\sqrt x+1}\\\quad =-(\sqrt x+1)+2-\dfrac{4}{\sqrt x+1}\\\quad =-\left[ (\sqrt x+1)+\dfrac{4}{\sqrt x+1}\right]+2$
Áp dụng BĐT Cô si với 2 số dương $\sqrt x+1;\dfrac{4}{\sqrt x+1}$
$\sqrt x+1+\dfrac{4}{\sqrt x+1}\ge 2\sqrt{(\sqrt x+1).\dfrac{4}{\sqrt x+1}}\\↔\sqrt x+1+\dfrac{4}{\sqrt x+1}\ge 4\\↔-\left[(\sqrt x+1)+\dfrac{4}{\sqrt x+1}\right]\le -4\\↔P\le -2\\→\max P=-2$
$→$ Dấu "=" xảy ra khi $\sqrt x+1=\dfrac{4}{\sqrt x+1}$
$↔(\sqrt x+1)^2=4$
mà $\sqrt x+1\ge 1>0$
$↔\sqrt x+1=2\\↔\sqrt x=1\\↔x=1(TM)$
Vậy $\max P=-2$ khi $x=1$
c/ ĐKXĐ: $x\ge 0$
$P\,=\dfrac{-2x-16}{\sqrt x+1}\\\quad =\dfrac{-2x-4\sqrt x-2+4\sqrt x+4-18}{\sqrt x+1}\\\quad =\dfrac{(-2x-4\sqrt x-2)+(4\sqrt x+4)-18}{\sqrt x+1}\\\quad =\dfrac{-2(x+2\sqrt x+1)+4(\sqrt x+1)-18}{\sqrt x+1}\\\quad =\dfrac{-2(\sqrt x+1)^2+4(\sqrt x+1)-18}{\sqrt x+1}\\\quad =-2(\sqrt x+1)+4-\dfrac{18}{\sqrt x+1}\\\quad =-\left[2(\sqrt x+1)+\dfrac{18}{\sqrt x+1}\right]+4$
Áp dụng BĐT Cô si với 2 số dương $2(\sqrt x+1);\dfrac{18}{\sqrt x+1}$
$2(\sqrt x+1)+\dfrac{18}{\sqrt x+1}\ge 2\sqrt{2(\sqrt x+1).\dfrac{18}{\sqrt x+1}}\\↔2(\sqrt x+1)+\dfrac{18}{\sqrt x+1}\ge 12\\↔-\left[2(\sqrt x+1)+\dfrac{18}{\sqrt x+1}\right]\le -12\\↔P\le -8\\→\max P=-8$
$→$ Dấu "=" xảy ra khi $2(\sqrt x+1)=\dfrac{18}{\sqrt x+1}$
$↔2(\sqrt x+1)^2=18\\↔(\sqrt x+1)^2=9$
mà $\sqrt x+1\ge 1>0$
$→\sqrt x+1=3\\↔\sqrt x=2\\↔x=4(TM)$
Vậy $\max P=-8$ khi $x=4$
d/ ĐKXĐ: $x\ge 0$
$P\,=\dfrac{-x-5}{\sqrt x+2}\\\quad =\dfrac{-x-4\sqrt x-4+4\sqrt x+8-9}{\sqrt x+2}\\\quad =\dfrac{-(x+4\sqrt x+4)+(4\sqrt x+8)-9}{\sqrt x+2}\\\quad =\dfrac{-(\sqrt x+2)^2+4(\sqrt x+2)-9}{\sqrt x+2}\\\quad =-(\sqrt x+2)+4-\dfrac{9}{\sqrt x+2}\\\quad =-\left[(\sqrt x+2)+\dfrac{9}{\sqrt x+2}\right]+4$
Áp dụng BĐT Cô si với 2 số dương $\sqrt x+2,\dfrac{9}{\sqrt x+2}$
$\sqrt x+2+\dfrac{9}{\sqrt x+2}\ge 2\sqrt{(\sqrt x+2).\dfrac{9}{\sqrt x+2}}\\↔\sqrt x+2+\dfrac{9}{\sqrt x+2}\ge 6\\↔-\left[(\sqrt x+2)+\dfrac{9}{\sqrt x+2}\right]\le -6\\↔P\le -2\\→\max P=-2$
$→$ Dấu "=" xảy ra khi $\sqrt x+2=\dfrac{9}{\sqrt x+2}$
$↔(\sqrt x+2)^2=9$
mà $\sqrt x+2\ge 2>0$
$→\sqrt x+2=3\\↔\sqrt x=1\\↔x=1(TM)$
Vậy $\max P=-2$ khi $x=1$
