Lời giải:
$(2x+3)^2+(2x+5)^2=4x^2+12x+9+4x^2+20x+25$

$=8x^2+32x+34$

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

=\(\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}\)

=\(\dfrac{5}{11}\times1+\dfrac{6}{11}\)

=\(\dfrac{11}{11}\)=1

\(\dfrac{5}{11}\cdot\dfrac{5}{7}+\dfrac{5}{11}\cdot\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=1\)

12x²  + 8x - 9x - 6 - 12x² + 30x + 2x -5 + 1= 31x - 10

\(:P=5x(x-3)(x+3)-(2x-3)^2+34x(x+2)-5(x+2)^3+25x-1\)

\(P=5x(x^2-9)-(4x^2-12x+9)+34x^2+68x-5(x^3+6x^2+12x+8)+25-1\)

\(P=5x^3-45x-4x^2+12x-9+34x^2+68x-5x^3-30x^2-60x-40+25-1\)

\(P=(5x^3-5x^3)+(34x^2-4x^2-30x^2)+(12x-45x++68x+25x-60x)-(9+1)\)

\(P=-10\)