[tex]2\left ( 5x-3 \right )\sqrt{x+1}+5\left ( x+1 \right )\sqrt{3-x}=3(5x+1)[/tex]
ĐKXĐ $-1\leq x\leq 3$
$2(5x+1)\sqrt{x+1} -8\sqrt{x+1}+(5x+1)\sqrt{3-x}+4\sqrt{3-x}=3(5x+1)$
$2(5x+1)\sqrt{x+1}-4(2\sqrt{x+1}-\sqrt{3-x})+(5x+1)\sqrt{3-x}=3(5x+1)$
$\left[\begin{matrix}x=\dfrac{-1}{5}\\2\sqrt{x+1}-\dfrac{4}{2\sqrt{x+1}+\sqrt{3-x}}+\sqrt{3-x}=3\end{matrix}\right.$
Đặt: $u=\sqrt{x+1}, v=\sqrt{3-x}\Rightarrow u^2+v^2=4$
Suy ra $2u-\dfrac{4}{2u+v}+v=3$
$\Rightarrow 3u^2+4uv=6u+3v$
$\Rightarrow 3u(u-2)=v(3-4u)$
$\Rightarrow 3u(u-2)=\sqrt{4-u^2}(3-4u)$
$\Rightarrow \left[\begin{matrix}u=2\\-3u\sqrt{2-u}=\sqrt{u+2}(3-4u)\end{matrix}\right.$
$\Rightarrow \left[\begin{matrix}x+1=4\\9u^2(2-u)=(u+2)(4u-3)^2\end{matrix}\right.$
$\Rightarrow \left[\begin{matrix}x=3\\25u^3-10u^2-39u+18=0\end{matrix}\right.$
Từ đây b giải tiếp nhé <3
Có gì khúc mắc b hỏi lại nhé <3
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
