ĐKXĐ: x>0; x<>9

\(A=\left(\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{4x}{x-9}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\)

\(=\dfrac{-x-6\sqrt{x}-9+x-6\sqrt{x}+9-4x}{x-9}:\dfrac{-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-4x-12\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}-2\right)}=\dfrac{4x}{\sqrt{x}-2}\)

|A|>-A

=>A>=0

=>4x>0

=>x>0 và x<>9

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

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

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

c: 2x-3căn x-5=0

=>2x-5căn x+2căn x-5=0

=>2căn x-5=0

=>x=25/4

Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)

2)

\(A=\dfrac{5\sqrt{a}-3}{\sqrt{a}-2}+\dfrac{3\sqrt{a}+1}{\sqrt{a}+2}-\dfrac{a^2+2\sqrt{a}+8}{a-4}\)

    \(=\dfrac{\left(5\sqrt{a}-3\right)\left(\sqrt{a}+2\right)+\left(3\sqrt{a}+1\right)\left(\sqrt{a}-2\right)-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

    \(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}+\sqrt{a}-2-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

    \(=\dfrac{-a^2+8a-16}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{-\left(a-4\right)^2}{a-4}=4-a\)

1: Ta có: \(\left\{{}\begin{matrix}3x-y=2m-1\\x+y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=5m+1\\x+y=3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=3m+2-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=\dfrac{12m+8-5m-1}{4}=\dfrac{7m+7}{4}\end{matrix}\right.\)

Ta có: \(x^2+2y^2=9\)

\(\Leftrightarrow\left(\dfrac{5m+1}{4}\right)^2+2\cdot\left(\dfrac{7m+7}{4}\right)^2=9\)

\(\Leftrightarrow\dfrac{25m^2+10m+1}{16}+\dfrac{2\cdot\left(49m^2+98m+49\right)}{16}=9\)

\(\Leftrightarrow25m^2+10m+1+98m^2+196m+98-144=0\)

\(\Leftrightarrow123m^2+206m-45=0\)

Đến đây bạn tự làm nhé, chỉ cần giải phương trình bậc hai bằng delta thôi

\(1,ĐKx\ge5\)

\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

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

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

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

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

Khi x=4-2căn 3 thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\sqrt{3}+3}{2}\)

a: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}+4}{x\sqrt{x}-3x+2\sqrt{x}}-\dfrac{3\sqrt{x}+3}{-x+\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}-6}{x-3\sqrt{x}}-\dfrac{x-2\sqrt{x}}{x-4\sqrt{x}+4}\right)+\sqrt{x}\)

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

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

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

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

=-1

a) \(P=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{x-9}.\dfrac{x-9}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)

b) \(P=\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

a: \(=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{2\sqrt{x}}\)

\(=\sqrt{x}\)