ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

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

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

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)

ai giúp mih trả lời với

a)Tại \(x=\frac{16}{9}\) ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{\frac{16}{9}}+1}{\sqrt{\frac{16}{9}}-1}=\frac{\frac{4}{3}+1}{\frac{4}{3}-1}=\frac{\frac{7}{3}}{\frac{1}{3}}=7\)

Tại \(x=\frac{25}{9}\) ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{\frac{25}{9}}+1}{\sqrt{\frac{25}{9}}-1}=\frac{\frac{5}{3}+1}{\frac{5}{3}-1}=\frac{\frac{8}{3}}{\frac{2}{3}}=4\)

b)Khi \(A=5\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}=5\)(*)

Đk:\(\sqrt{x}-1\ne0\Rightarrow x\ne1;\sqrt{x}\ge0\Rightarrow x\ge0\)

Đặt \(\sqrt{x}+1=t\left(t\ge0\right)\),(*) trở thành

\(\frac{t}{t-2}=5\Rightarrow t=5\left(t-2\right)\)

\(\Rightarrow t=5t-10\)

\(\Rightarrow2t=5\Rightarrow t=\frac{5}{2}\)(thỏa mãn)

\(t=\frac{5}{2}\Rightarrow\sqrt{x}+1=\frac{5}{2}\)

\(\Rightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow\sqrt{x^2}=\left(\frac{3}{2}\right)^2\Leftrightarrow x=\frac{9}{4}\)(thỏa mãn)

Vậy \(x=\frac{9}{4}\)

a) Thay \(x=\frac{16}{9}\) vào biểu thức ta có:

\(A=\frac{\sqrt{\frac{16}{9}}+1}{\sqrt{\frac{16}{9}}-1}=\frac{\frac{4}{3}+1}{\frac{4}{3}-1}=\frac{\frac{7}{3}}{\frac{1}{3}}=7\)

Vậy \(A=7\)

Thay \(x=\frac{25}{9}\) vào biểu thức ta có:

\(A=\frac{\sqrt{\frac{25}{9}}+1}{\sqrt{\frac{25}{9}}-1}=\frac{\frac{5}{3}+1}{\frac{5}{3}-1}=\frac{\frac{8}{3}}{\frac{2}{3}}=4\)

Vậy \(A=4\)

a)\(\frac{\left(x-1\right)}{\sqrt{x}}\)

b) để P>0\(\Rightarrow\)\(\frac{\left(x-1\right)}{\sqrt{x}}>0\)

do \(\sqrt{x}>0\Rightarrow x-1>0\)

\(\Leftrightarrow x>1\)

c)P=\(\frac{8}{3}\)

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