Đk: $-5Pt $\Leftrightarrow 3(5-x-2)\sqrt{5+x}+3(5+x-2)\sqrt{5-x}=4\sqrt{25-x^2}$$\Leftrightarrow 3\sqrt{25-x^2}(\sqrt{5-x}+\sqrt{5+x})-6(\sqrt{5-x}+\sqrt{5+x})=4\sqrt{25-x^2}$$\Leftrightarrow 3(\sqrt{5-x}+\sqrt{5+x})(\sqrt{25-x^2}-2)=4\sqrt{25-x^2}$Đặt $t=\sqrt{5-x}+\sqrt{5+x}\geq 0\Rightarrow \frac{t^2-10}{2}=\sqrt{25-x^2}(*)$Pt trở thành: $3t^3-4t^2-42t+40=0\Leftrightarrow (t-4)(3t^2+8t-10)=0\Leftrightarrow \left[ {\begin{matrix} t=4\\ 3t^2+8t-10=0 \end{matrix}} \right.$Thế vào $(*)$ rồi tìm nghiệm nhé!
Đk: $-5<x<5$Pt $\Leftrightarrow 3(5-x-2)\sqrt{5+x}+3(5+x-2)\sqrt{5-x}=4\sqrt{25-x^2}$$\Leftrightarrow 3\sqrt{25-x^2}(\sqrt{5-x}+\sqrt{5+x})-6(\sqrt{5-x}+\sqrt{5+x})=4\sqrt{25-x^2}$$\Leftrightarrow 3(\sqrt{5-x}+\sqrt{5+x})(\sqrt{25-x^2}-2)=4\sqrt{25-x^2}$Đặt $t=\sqrt{5-x}+\sqrt{5+x}\Rightarrow \frac{t^2-10}{2}=\sqrt{25-x^2}(*)$Pt trở thành: $3t^3-4t^2-42t+40=0\Leftrightarrow (t-4)(3t^2+8t-10)=0\Leftrightarrow \left[ {\begin{matrix} t=4\\ 3t^2+8t-10=0 \end{matrix}} \right.$Thế vào $(*)$ rồi tìm nghiệm nhé!
Đk: $-5Pt $\Leftrightarrow 3(5-x-2)\sqrt{5+x}+3(5+x-2)\sqrt{5-x}=4\sqrt{25-x^2}$$\Leftrightarrow 3\sqrt{25-x^2}(\sqrt{5-x}+\sqrt{5+x})-6(\sqrt{5-x}+\sqrt{5+x})=4\sqrt{25-x^2}$$\Leftrightarrow 3(\sqrt{5-x}+\sqrt{5+x})(\sqrt{25-x^2}-2)=4\sqrt{25-x^2}$Đặt $t=\sqrt{5-x}+\sqrt{5+x}
\geq 0\Rightarrow \frac{t^2-10}{2}=\sqrt{25-x^2}(*)$Pt trở thành: $3t^3-4t^2-42t+40=0\Leftrightarrow (t-4)(3t^2+8t-10)=0\Leftrightarrow \left[ {\begin{matrix} t=4\\ 3t^2+8t-10=0 \end{matrix}} \right.$Thế vào $(*)$ rồi tìm nghiệm nhé!