\(P=\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

\(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}-2}{x-4}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(P=\left(\dfrac{-6}{x-4}\right):\left(\dfrac{6}{\sqrt{x}+2}\right)=\dfrac{-1}{\sqrt{x}-2}\)

mik cần gấp ạ^^

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

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

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)

b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)

=>x<1

=>x<1

1: Ta có: \(A=\left(\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}\right)-\dfrac{\sqrt{x}+6}{4-x}\)

\(=\dfrac{2\sqrt{x}+x+2\sqrt{x}-x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\dfrac{\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{-3}{\sqrt{x}+2}\)

a: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

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

\(=x+\sqrt{x}-2\sqrt{x}-1+1\)

\(=x-\sqrt{x}\)

b: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)

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

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

ĐKXĐ: \(x\ge0;x\ne4\)

\(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)

\(=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

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

Khi \(x=9+4\sqrt{5}\)

Ta có: \(4+4\sqrt{5}+5=2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(2+\sqrt{5}\right)^2\)

\(\Rightarrow\sqrt{x}=2+\sqrt{5}\)

\(\Rightarrow P=\dfrac{\left(2+\sqrt{5}\right)^2+1}{2+\sqrt{5}-2}=\dfrac{9+4\sqrt{5}+1}{\sqrt{5}}=\dfrac{10+4\sqrt{5}}{\sqrt{5}}=4+2\sqrt{5}\)

Vậy \(P=4+2\sqrt{5}\) khi \(x=9+4\sqrt{5}\).

\(D=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{x-4}\)

\(=\dfrac{x\sqrt{x}+2x+2}{x-4}\)

Khi x=9+4căn 5 thì \(D=\dfrac{\left(9+4\sqrt{5}\right)\left(\sqrt{5}+2\right)+2\sqrt{5}+4+2}{\sqrt{5}-2}\)

\(=\dfrac{9\sqrt{5}+18+20+8\sqrt{5}+2\sqrt{5}+6}{\sqrt{5}-2}\)

=(44+19căn 5)*(căn 5+2)

=44căn 5+88+95+38căn 5

=82căn 5+183

Lời giải:
\(A=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{5}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{\sqrt{x}+3}{(\sqrt{x}-2)(\sqrt{x}+3)}\)

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