ĐK:x>0,x\(\ne\)9

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right)\div\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right)\div\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\div\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\dfrac{-3\left(\sqrt{x}+3\right).\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\left(2\sqrt{x}+4\right)}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

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

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

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

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

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

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

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

\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}+\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{2}\right)}{-\sqrt{x}}\)

\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)

CHÚC BẠN HỌC TỐT NHÉ

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

bn ơi! sai đề thì pải

a: \(C=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{1}{x-\sqrt{x}+1}\)

\(=\dfrac{x-\sqrt{x}+1-3+\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\dfrac{x-1}{x\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)

b: Để C<1 thì C-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1-x+\sqrt{x}-1}{x-\sqrt{x}+1}< 0\)

=>\(-x+2\sqrt{x}-2< 0\)(luôn đúng)

Đặt \(\sqrt[3]{2}=a\)

\(A+\dfrac{a+a^2+a^3}{a^2+a+1}=\dfrac{a\left(a^2+a+1\right)}{a^2+a+1}=a=\sqrt[3]{2}\)