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) x=16/9 => A = 6

    x=25/9 => A = 3

b) A = 5 (=) x = 35/25

k cho mik nha

A = căn x +1 trên căn x -1

A = căn x - 1 + 3 trên căn x - 1

A = 1 cộng vs 3 trên căn x - 1

thay x = 16/9

A = 1+ vs 3 trên căn 16/9 -1

A = 1 + vs 3 trên 4/3 - 1

A = 1+ vs 3 trên 1/3

A = 1+ vs 9

A= 10

tương tự vs x =25/9

A=5

=> 5 =1 + vs 3 trên căn x -1

4 = 3 trên căn x -1 

căn x-1 = 3/4

căn x = 7/3

x = 49/9

đúng đấy

bạn DUY ĐẠT ơi , tại sao căn x - 1 + 3 lại = đc căn x - 1 . Phải là căn x - 1 +2 nhỉ

Ai đó giúp mk vs ! 

Em gửi ảnh ạ !

Em gửi ảnh trên ạ !!!!!

a, Ta có \(x=49\Rightarrow\sqrt{x}=7\)

Thay vào biểu thức A ta được : 

\(A=\frac{7.4}{7-1}=\frac{28}{6}=\frac{14}{3}\)

b, Với \(x\ge0;x\ne1\)

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

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

\(=\frac{3\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

x=\(24-16\sqrt{2}=4^2-2.4.\sqrt{8}+\left(2\sqrt{2}\right)^2=\left(4-2\sqrt{2}\right)^2\)

a) \(P=\frac{3}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}-5}{x-1}\)

\(P=\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{x-1}\)

\(P=\frac{3\sqrt{x}-3-\sqrt{x}-1}{x-1}-\frac{\sqrt{x}-5}{x-1}\)

\(P=\frac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{x-1}\)

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

vay \(P=\frac{\sqrt{x}+1}{x-1}\)

b)  thay vao P ta duoc:

\(P=\frac{\sqrt{24-16\sqrt{2}}+1}{24-16\sqrt{2}-1}\)

\(P=\frac{\sqrt{\left(2\sqrt{2}\right)^2-2.2.4\sqrt{2}+4^2}+1}{\left(2\sqrt{2}\right)^2-2.2.4\sqrt{2}+4^2-1}\)

\(P=\frac{\sqrt{\left(2\sqrt{2}-4\right)^2}+1}{\left(2\sqrt{2}-4\right)^2-1^2}\)

\(P=\frac{2\sqrt{2}-4+1}{\left(2\sqrt{2}-4-1\right)\left(2\sqrt{2}-4+1\right)}\)

\(P=\frac{2\sqrt{2}-3}{\left(2\sqrt{2}-5\right)\left(2\sqrt{2}-3\right)}\)

\(P=\frac{1}{2\sqrt{2}-5}\)

vay \(P=\frac{1}{2\sqrt{2}-5}\)

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 \(x=\frac{9}{4}\)