`a)`\(x^2-7x+\sqrt{x^2-7x+8}=12\)
\(ĐK:x^2-7x+8\ge0\)
Đặt \(x^2-7x=a\)
\(\Leftrightarrow a+\sqrt{a+8}=12\)
\(\Leftrightarrow\sqrt{a+8}=12-a\)
\(\Leftrightarrow a+8=144-24a+a^2\)
\(\Leftrightarrow a^2-25a+136=0\)
\(\Delta=\left(-25\right)^2-4.136=81>0\)
\(\rightarrow\left\{{}\begin{matrix}x=\dfrac{25+\sqrt{81}}{2}=17\left(tm\right)\\x=\dfrac{25-\sqrt{81}}{2}=8\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{17;8\right\}\)
`c)`\(\sqrt{x-3}+\sqrt{5-x}=x^2-8x+18\)
\(ĐK:3\le x\le5\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(\left(\sqrt{x-3}+\sqrt{5-x}\right)^2\le2\left(x-3+5-x\right)\)
\(\Leftrightarrow\left(\sqrt{x-3}+\sqrt{5-x}\right)^2\le4\)
\(\Leftrightarrow\sqrt{x-3}+\sqrt{5-x}\le2\)
Dấu "=" xảy ra khi \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{5-x}}\)
\(\Leftrightarrow x-3=5-x\)
\(\Leftrightarrow x=4\)
\(x^2-8x+18=\left(x^2-8x+16\right)+2=\left(x-4\right)^2+2\ge2\)
Dấu"=" xảy ra khi \(x=4\)
`=>` pt xảy ra khi và chỉ khi `x=4` `(tm)`
Vậy \(S=\left\{4\right\}\)
