Gợi ý

ĐKXĐ: ....

Do x=0 không phải là nghiệm nên chia cả hai vế cho x^2 có

\(\sqrt{2+\frac{5}{x}+\frac{3}{x^2}}=4-\frac{5}{x}-\frac{3}{x^2}\)(1) Đặt \(\sqrt{\frac{5}{x}+\frac{3}{x^2}+2}=y\Rightarrow y\ge0\)và \(\frac{5}{x}+\frac{3}{x^2}=y^2-2\)

Khi đó \(\left(1\right)\Leftrightarrow y=4-y^2+2\)Sau khi tìm được y thì thế vào tìm x ,  rồi đối chiếu ĐKXĐ và trả lời

   KL : ...

p/s: Amasterasu ăn nói cho hẳn hoi nhắc bn thêm lần nx

Rút gọn thôi à ._.

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

ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

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

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

a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)

b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)

\(ĐK:x\ge-1\)

\(x^2+x+12\sqrt{x+1}=36\Leftrightarrow\left(x^2+x-12\right)+\left(12\sqrt{x+1}-24\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)+12.\frac{x-3}{\sqrt{x+1}+2}=0\Leftrightarrow\left(x-3\right)\left(x+4+\frac{12}{\sqrt{x+1}+2}\right)=0\)

Dễ thấy \(x+4+\frac{12}{\sqrt{x+1}+2}>0\forall x\ge-1\)nên x - 3 = 0 hay x = 3 (tm)

Vậy nghiệm duy nhất của phương trình là 3

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

ĐKXĐ : x > 1

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

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

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

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

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

Để A = 9/2

=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )

<=> 2x = 9( √x - 1 )

<=> 2x = 9√x - 9

<=> 2x + 9 = 9√x (1)

Bình phương hai vế

(1) <=> 4x² + 36x + 81 = 81x

     <=> 4x² + 36x + 81 - 81x = 0

     <=> 4x² - 45x + 81 = 0

     <=> 4x² - 36x - 9x + 81 = 0

     <=> 4x( x - 9 ) - 9( x - 9 ) = 0

     <=> ( x - 9 )( 4x - 9 ) = 0

     <=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )