\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13,đkxđ:-1\le x\le7,\Leftrightarrow\left(\sqrt{7-x}+\sqrt{x+1}\right)^2=\left(x^2-6x+13\right)^2\Leftrightarrow7-x+x+1+2\sqrt{\left(7-x\right)\left(x+1\right)}=\left(x^2-6x+13\right)\left(x^2-6x+13\right)\Leftrightarrow8+2\sqrt{7x+8-x^2-x}=x^4-6x^3+13x^2-6x^3+36x^2-78x+13x^2-78x+169\Leftrightarrow8+2\sqrt{-x^2+6x+8}=x^4-12x^3+62x^2-120x+169\Leftrightarrow Bírồi:< \)
\(Chot=7-x\Rightarrow x=7-t\Rightarrow\sqrt{7-x}=\sqrt{7-7+t}=\sqrt{t}và\sqrt{x+1}=\sqrt{7-t+1}=\sqrt{8-t}vàx^2-6x+13=\left(7-t\right)^2-6\left(7-t\right)+13,tacópt:\sqrt{t}+\sqrt{8-t}=49-14t+t^2-42+6t+13\Leftrightarrow\sqrt{t}+\sqrt{8-t}=t^2-8t+20=t^2-2.4.t+16+4=\left(t-4\right)^2+4\Leftrightarrow\left(\sqrt{t}+\sqrt{8-t}\right)^2=\left[\left(t-4\right)^2+4\right]^2\Leftrightarrow t-t+8+2\sqrt{8t-t^2}=...\left(bítiếp\right)\)
\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13\left(đk:-1\le x\le7\right)\)
Với a,b>0 ta AD BĐT: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\) (tự CM nha 
).Dấu "=" xảy ra<=>a=b (1)
AD bđt (1) có:
\(\sqrt{7-x}+\sqrt{x+1}\le2\sqrt{\frac{7-x+x+1}{2}}\)
\(\le2\sqrt{4}\) =4 (*)
Có x2-6x+13=(x-3)2+4 \(\ge4\) (**)
Từ (*),(**) => Dấu bằng xảy ra \(< =>\left\{{}\begin{matrix}7-x=x+1\\x-3=0\end{matrix}\right.\) \(< =>\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)\(< =>x=3\)(tm điều kiện của x)
Vậy x=3
