ĐK \(\orbr{\begin{cases}x>2\\x\le-2\end{cases}}\)

Đặt \(\sqrt{\frac{x+2}{x-2}}=t\Rightarrow x+2=t^2\left(x-2\right)\)

Vậy thì phương trình trở thành \(t^2\left(x-2\right)^2+4\left(x-2\right)t+3=0\)

\(\Leftrightarrow\left[t\left(x-2\right)+1\right]\left[t\left(x-2\right)+3\right]=0\)

Với \(t\left(x-2\right)+1=0\Leftrightarrow\sqrt{\frac{x+2}{x-2}}\left(x-2\right)+1=0\)

Để pt có nghiệm thì \(x-2< 0\) , khi đó \(-\sqrt{\frac{x+2}{x-2}\left(x-2\right)^2}+1=0\Leftrightarrow-\sqrt{x^2-4}+1=0\)

\(\Leftrightarrow x^2-4=1\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\left(l\right)\\x=-\sqrt{5}\left(n\right)\end{cases}}\)

Với \(t\left(x-2\right)+3=0\Leftrightarrow-\sqrt{x^2-4}+3=0\)

\(\Leftrightarrow x^2-4=9\Leftrightarrow\orbr{\begin{cases}x=\sqrt{13}\left(l\right)\\x=-\sqrt{13}\left(n\right)\end{cases}}\)

Vậy pt có tập nghiệm \(S=\left\{-\sqrt{13};-\sqrt{5}\right\}\)

@Azue help me

help me

mọi người ưi giúp tui giải câu a thui nha tui giải đc câu b ròi làm ơn nhanh giúp thanks nhìu nhìu

đặt \(\sqrt{x+1}=a;\sqrt{x-2}=b\left(a,b\ge0\right)\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x^2-x-2}=ab\\3=a^2-b^2\end{cases}}\)

PT đã cho trở thành : \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\Rightarrow\left(a-b\right)\left(a+b-ab-1\right)=0\)

\(\Rightarrow\left(a-b\right)\left(a-1\right)\left(1-b\right)=0\)

từ đó giải ra được x

Bài 1 :

a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)

= \(2\left|a-3\right|+2\left|a+2\right|\)

\(=2.\left(-a+3\right)+2\left(-a-2\right)\)

b) có sai đề ko ?

c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)

a) Ta có: \(P=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)

\(=\left(\dfrac{2}{\sqrt{xy}}+\dfrac{1}{x}+\dfrac{1}{y}\right):\dfrac{x\sqrt{x}+y\sqrt{x}+x\sqrt{y}+y\sqrt{y}}{x\sqrt{xy}+y\sqrt{xy}}\)

\(=\left(\dfrac{x+2\sqrt{xy}+y}{xy}\right):\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

a) Đk:\(x>0;y>0\)

\(P=\left[\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}.\sqrt{y}}.\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{x\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{xy}+y\sqrt{xy}}\)

\(=\left[\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\right]:\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)

\(=\dfrac{2\sqrt{xy}+x+y}{xy}:\dfrac{\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}.\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

b) \(xy=16\Leftrightarrow x=\dfrac{16}{y}\)

\(P=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=\dfrac{1}{\sqrt{\dfrac{16}{y}}}+\dfrac{1}{\sqrt{y}}=\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\)

Áp dụng AM-GM có:

\(\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\ge2\sqrt{\dfrac{\sqrt{y}}{4}.\dfrac{1}{\sqrt{y}}}=1\)

\(\Rightarrow P\ge1\)

Dấu "=" xảy ra khi \(y=4\Rightarrow x=4\)

Vậy x=y=4 thì P đạt GTNN là 1

Ta có: \(\sqrt{2x^2+7x+10}+\sqrt{2x^2+x+4}=3x+3\)

\(\Rightarrow\sqrt{2x^2+7x+10}+\sqrt{2x^2+x+4}-3x-3=0\)

\(\Rightarrow\sqrt{2x^2+7x+10}-7+\sqrt{2x^2+x+4}-5-3x+9=0\)

\(\Rightarrow\frac{2x^2+7x+10-49}{\sqrt{2x^2+7x+10}+7}+\frac{2x^2+x+4-25}{\sqrt{2x^2+x+4}+5}-3\left(x-3\right)=0\)

\(\Rightarrow\frac{\left(x-3\right)\left(2x+13\right)}{\sqrt{2x^2+7x+10}+7}+\frac{\left(x-3\right)\left(2x+7\right)}{\sqrt{2x^2+x+4}+5}-3\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(\frac{2x+13}{\sqrt{2x^2+7x+10}+7}+\frac{2x+7}{\sqrt{2x^2+x+4}+5}-3\right)=0\)

mà \(\frac{2x+13}{\sqrt{2x^2+7x+10}+7}+\frac{2x+7}{\sqrt{2x^2+x+4}}-3< 0\)

=> x - 3 = 0 => x = 3

                                                              Vậy x = 3

dân dương ơi bài này dễ mà nhân liên hợp là ok