Question

Giải phương trình:

\(\sqrt{x-4}+\sqrt{6-x}=x^2-10x+27\)

\(\sqrt{x-2}+\sqrt{4-x}=x^2-6x+11\)

Answer 1

bài 1 :điều kiện\(4\le x\le6\) 

 ta có \(VT=\left(\sqrt{x-4}+\sqrt{6-x}\right)\le\sqrt{2\left(x-4+6-x\right)}=\sqrt{2\cdot2}=2\)

\(VP=x^2-10x+27=x^2-10x+25+2=\left(x-5\right)^2+2\ge2\)

\(\Rightarrow VT=VP=2\Leftrightarrow x=5\)(t/m)

bài 2 :điều kiện : \(2\le x\le4\)

ta có \(VT=\left(\sqrt{x-2}+\sqrt{4-x}\right)\le\sqrt{2\left(x-2+4-x\right)}=2\)

\(VP=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)

\(\Rightarrow VT=VP=2\Leftrightarrow x=3\)(t/m)