ĐK: \(y\ne0,xy\ge0\).

\(4x^2+9y^2=16xy\)

Chia cả hai vế cho \(y^2\)ta được: 

\(4\left(\frac{x}{y}\right)^2+9=\frac{16x}{y}\)

\(\Leftrightarrow\frac{x}{y}=\frac{4\pm\sqrt{7}}{2}\)

Với \(y>0\)thì \(x\ge0\)

\(P=\frac{\sqrt{xy}+\sqrt{y^2}}{y}-\sqrt{\frac{x}{y}}=\frac{\sqrt{x}\sqrt{y}+y}{y}-\sqrt{\frac{x}{y}}=\sqrt{\frac{x}{y}}+1-\sqrt{\frac{x}{y}}=1\)

Với \(y< 0\)thì \(x\le0\):

\(P=\frac{\sqrt{xy}+\sqrt{y^2}}{y}-\sqrt{\frac{x}{y}}=\frac{\sqrt{-x}\sqrt{-y}-y}{y}-\sqrt{\frac{x}{y}}=-\sqrt{\frac{x}{y}}-1-\sqrt{\frac{x}{y}}=-2\sqrt{\frac{x}{y}}-1\)

\(=-2\sqrt{\frac{4\pm\sqrt{7}}{2}}-1=-\left(1\pm\sqrt{7}\right)-1=-2\pm\sqrt{7}\)

a) \(2x^2-6=0\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt{3}\).

b) \(x^2-2\sqrt{5}x=0\)

\(\Leftrightarrow x\left(x-2\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\sqrt{5}\end{cases}}\)

c) \(x^2+6x+9=3\)

\(\Leftrightarrow\left(x+3\right)^2=3\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{3}\\x+3=-\sqrt{3}\end{cases}}\)

\(\Leftrightarrow x=-3\pm\sqrt{3}\)

Bài 3 : 

a, \(A=\frac{3+\sqrt{5}}{\sqrt{5}+2}+\frac{\sqrt{5}}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\frac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)

\(=\sqrt{5}-1+\frac{5+\sqrt{5}-9\sqrt{5}+15}{4}=\sqrt{5}-1+\frac{20-7\sqrt{5}}{4}\)

\(=\frac{4\sqrt{5}-4+20-7\sqrt{5}}{4}=\frac{-3\sqrt{5}+16}{4}\)

b, Với x >  0 

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

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

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

bổ sung đề : mình ko biết đúng ko nhưng ko phải đăng lại giải lại :)) 

a,  \(A=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)ĐK : \(x>0\)

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

b, Với \(A=\frac{9}{2}\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}=\frac{9}{2}\)

\(\Rightarrow2x+2\sqrt{x}+2=9\sqrt{x}\Leftrightarrow2x-7\sqrt{x}+2=0\)

\(\Leftrightarrow x=\sqrt{\frac{7\pm\sqrt{33}}{4}}\) kết hợp với đk vậy : \(x=\sqrt{\frac{7+\sqrt{33}}{4}}\)

Đáp án :

- Bình phương lê nha bn

B = \(\sqrt{4\sqrt{6}+8\sqrt{3}+4\sqrt{2}+18}=\sqrt{12+2+4+4\sqrt{6}+8\sqrt{3}+4\sqrt{2}}\)

= \(\sqrt{\left(2\sqrt{3}+\sqrt{2}+2\right)^2}=2\sqrt{3}+\sqrt{2}+2\)

\(B=\sqrt{18+4\sqrt{6}+8\sqrt{3}+4\sqrt{2}}\)

\(B=\sqrt{12+2+4+4\sqrt{6}+8\sqrt{3}+4\sqrt{2}}\)

\(B=\sqrt{\left(2\sqrt{3}\right)^2+\left(\sqrt{2}^2\right)+2^2+2.2\sqrt{3}\sqrt{2}+2.2\sqrt{3}.2+2.\sqrt{2}.2}\)

\(B=\sqrt{\left(2\sqrt{3}+\sqrt{2}+2\right)^2}\)

\(B=2\sqrt{3}+\sqrt{2}+2\)

\(\sqrt{3}< 2;\sqrt{3}>1\)

\(\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\)

\(2-\sqrt{3}+\sqrt{3}-1\)

\(=1\)

\(b,\sqrt{5}< \sqrt{9}=3\)

\(\left|\sqrt{5}-3\right|-2\sqrt{5}+2\)

\(3-\sqrt{5}-2\sqrt{5}+2\)

\(5-3\sqrt{5}\)

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

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

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

Thay \(x=\frac{a^2+b^2}{2ab}\)vào A, ta được : 

\(A=\frac{\sqrt{\frac{a^2+b^2}{2ab}+1}+\sqrt{\frac{a^2+b^2}{2ab}-1}}{\sqrt{\frac{a^2+b^2}{2ab}+1}-\sqrt{\frac{a^2+b^2}{2ab}-1}}\)

\(A=\frac{\sqrt{\frac{\left(a+b\right)^2}{2ab}}+\sqrt{\frac{\left(b-a\right)^2}{2ab}}}{\sqrt{\frac{\left(a+b\right)^2}{2ab}}-\sqrt{\frac{\left(b-a\right)^2}{2ab}}}\)

\(A=\frac{a+b\sqrt{\frac{1}{2ab}}+\left(b-a\right)\sqrt{\frac{1}{2ab}}}{a+b\sqrt{\frac{1}{2ab}}-\left(b-a\right)\sqrt{\frac{1}{2ab}}}\)

\(A=\frac{a+b+b-a}{a+b-b+a}\)

\(A=\frac{2b}{2a}\)

\(A=\frac{b}{a}\)

                            Ps : Nhớ k cho tui nhó, tui đã rất cố gắng rồi đấy. :)) K để lần sau có j tui giải giúp cho :)))

                                                                                                                                         # Aeri #