\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{8}{5}< -6\\-6< x\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< -6+\dfrac{8}{5}\\x>-6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< -\dfrac{22}{5}\\x>-6\end{matrix}\right.\\ \Rightarrow-6< x< -\dfrac{22}{5}\)

\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\\ =\left(\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{x-4}\right):\left(\dfrac{\sqrt{x}+2+3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right)\\ =\left(\dfrac{4x-8\sqrt{x}-8x}{x-4}\right)\times\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+2+3\sqrt{x}-6}\right)\\ =\left(\dfrac{-8\sqrt{x}-4x}{x-4}\right)\times\left(\dfrac{\sqrt{x}-2}{4\sqrt{x}-4}\right)\\ =\left(\dfrac{-4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-2\right)}\right)\times\left(\dfrac{\sqrt{x}-2}{4\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{-4\sqrt{x}}{4\left(\sqrt{x}-1\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

b, \(P=-4\\ \Rightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-1}=-4\\ \Leftrightarrow-\sqrt{x}=-4\sqrt{x}+4\\ \Leftrightarrow-\sqrt{x}+4\sqrt{x}=4\\ \Leftrightarrow\sqrt{x}=\dfrac{4}{3}\\ \Rightarrow x=\dfrac{16}{9}\left(t/mđk\right)\)

Cho mình hỏi 1 vấn đề ngoài lề một xíu : Làm sao để có coin ạ 

Cho hình chóp \(SABCD\) gọi \(H,K\) lần lượt là trọng tâm \(\Delta SAB,\Delta SBC\) . M là trung điểm \(AC,I\in SM\) sao cho \(SI>SM\) .Tìm giao tuyến \(a,\left(IHK\right);\left(ABC\right)\\ b,\left(IHM\right);\left(SBC\right)\)

\(a,C=A+B\\ =4x^2+3y^2-5xy+3x^2+2y^2+2x^2y^2\\ =\left(4x^2+3x^2\right)+\left(3y^2+2y^2\right)-5xy+2x^2y^2\\ =7x^2+5y^{^2}-5xy+2x^2y^2\\ b,C+A=B\\ =>C=B-A\\ =\left(3x^2+2y^2+2x^2y^2\right)-\left(4x^2+3y^2-5xy\right)\\ =3x^2+2y^2+2x^2y^2-4x^2-3y^2+5xy\\ =\left(3x^2-4x^2\right)+\left(2y^2-3y^2\right)+2x^2y^2+5xy\\ =-x^2-y^2+2x^2y^2+5xy\)