Câu 2.$\sqrt{x+1}+\sqrt{4-x}+\sqrt{(x+1)(4-x)}=5$Đặt $t=\sqrt{x+1}+\sqrt{4-x}> 0$(Sử dụng BĐT BCS để tìm max t nhé)$\Rightarrow t^2=5+2\sqrt{(x+1)(4-x)}$.Khi đó Pt $\Leftrightarrow t+\frac{t^2-5}{2}=5$
Câu 2.$\sqrt{x+1}+\sqrt{4-x}+\sqrt{(x+1)(4-x)}=5$Đặt $t=\sqrt{x+1}+\sqrt{4-x}\geq 0\Rightarrow t^2=5+2\sqrt{(x+1)(4-x)}$.Khi đó Pt $\Leftrightarrow t+\frac{t^2-5}{2}=5$
Câu 2.$\sqrt{x+1}+\sqrt{4-x}+\sqrt{(x+1)(4-x)}=5$Đặt $t=\sqrt{x+1}+\sqrt{4-x}
&g
t; 0
$(Sử dụng BĐT BCS để tìm max t nhé)$\Rightarrow t^2=5+2\sqrt{(x+1)(4-x)}$.Khi đó Pt $\Leftrightarrow t+\frac{t^2-5}{2}=5$