Chủ đề này có 2 trả lời

[thinhnarutop]

$x\sqrt{3x+2}+\sqrt{4-x}=\sqrt{2(x^{2}+1)(x+3)}$

[An Infinitesimal]

$x\sqrt{3x+2}+\sqrt{4-x}=\sqrt{2(x^{2}+1)(x+3)}$

 

Bụp một bài bằng BĐT BCS là xong!

 

\[x\sqrt{3x+2}+\sqrt{4-x} \le \sqrt{\left(x^2+1\right)\left(2x+6\right)}= VP.\]

 

Xin dừng lại ở đây!

[huykinhcan99]

Ta có

\begin{align*} &\phantom{\iff ~} x\sqrt{3x+2}+\sqrt{4-x}=\sqrt{2(x^{2}+1)(x+3)} \\ &\iff x^2(3x+2)+4-x+2x\sqrt{(3x+2)(4-x)}=2(x^2+1)(x+3) \\ &\iff -x^3+4x^2+3x+2 -2x\sqrt{(3x+2)(4-x)}=0 \\ &\iff x^2(4-x)+(3x+2)-2x\sqrt{(3x+2)(4-x)}=0 \\&\iff \left(x\sqrt{4-x}-\sqrt{3x+2}\right)^2=0 \\ &\iff x\sqrt{4-x}=\sqrt{3x+2} \\ &\iff \left\{ \begin{array}{l} x^2(4-x)=3x+2 \\ x\geqslant 0\end{array} \right. \\ & \iff  \left\{ \begin{array}{l} x^3-4x+3x+2=0 \\ x\geqslant 0\end{array} \right. \\ &\iff  \left\{ \begin{array}{l}  \left[ \begin{array}{l} x=1+\sqrt{2} \\ x=1-\sqrt{2} \\ x=2\end{array} \right. \\ x\geqslant 0\end{array} \right. \\ & \iff  \left[ \begin{array}{l} x=1+\sqrt{2} \\ x=2\end{array} \right. \end{align*}