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\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)
\(A=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
\(minA=-56\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(B=4x-x^2+1=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)
\(maxB=5\Leftrightarrow x=2\)
ta có |a|+|b|\(\ge\text{|a+b|}\) dấu bằng xảy ra <=>a>0 và b>0
D=|x2+x+3|+|x2+x-6|=|x2+x+3|+|-x2-x+6|>|x2+x+3-x2-x+6|=9
minD=9<=>x2+x+3>0 dúng
x2+x-6>0 roi tu tinh x
\(D=\left|x^2+x+3\right|+\left|-x^2-x+6\right|\ge\left|x^2+x+3-x^2-x+6\right|=\left|9\right|=9\)
Vậy D min = 9