\(P=\dfrac{\left(\sqrt{a+1}+1\right)\left(\sqrt{a+1}+2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}+\dfrac{2\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}-\dfrac{2+5\sqrt{a+1}}{a-3}\)

\(P=\dfrac{a+3+3\sqrt{a+1}}{a-3}+\dfrac{2a+2-4\sqrt{a+1}}{a-3}-\dfrac{2+5\sqrt{a+1}}{a-3}\)

\(P=\dfrac{a+3+3\sqrt{a+1}+2a+2-4\sqrt{a+1}-2-5\sqrt{a+1}}{a-3}\)

\(P=\dfrac{3a+3-6\sqrt{a+1}}{a-3}\)

Có thể dừng ở đây hoặc nếu thích thì làm tiếp như sau (chưa chắc gọn hơn):

\(P=\dfrac{3\left(a+1\right)-6\sqrt{a+1}}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}=\dfrac{3\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}\)

\(P=\dfrac{3\sqrt{a+1}}{\sqrt{a+1}-2}\)

a + b + c = 0 => c = -a - b ; b= -a - c ; a =  - b - c 

Thay vào Q ta có :

\(Q=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{a^2+c^2-\left(a+c\right)^2}\)

\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{c^2+a^2-c^2-a^2-2ac}\)

\(Q=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{c+a+b}{-2abc}=0\)

\(X=\sqrt{a^2+1+\left(1-\frac{1}{a+1}\right)^2}+\frac{a}{a+1}\)

\(=\sqrt{a^2+1+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{\frac{\left(a^2+1\right)\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1\)