\(a\text{) ĐK: }15\le x\le97\)
Đặt \(a=\sqrt[4]{97-x};\text{ }b=\sqrt[4]{x-15}\text{ }\left(a;b\ge0\right)\)
Thì \(a^4+b^4=97-x+x-15=82\text{ (1)}\)
Mặt khác, pt đã cho thành \(a+b=4\Leftrightarrow b=4-a,\text{ thay vào (1) ta được: }\)
\(a^4+\left(4-a\right)^4=82\)
Đặt \(a-2=b;\text{ }b\ge-2\)
Pt trở thành \(\left(b+2\right)^4+\left(b-2\right)^4=82\Leftrightarrow b^4+24b^2-25=0\)
\(\Leftrightarrow\left(b^2-1\right)\left(b^2+25\right)=0\Leftrightarrow b^2=1\Leftrightarrow b=\pm1\)
\(+b=1\text{ thì }a=b+2=3\Rightarrow\sqrt[4]{97-x}=3\Leftrightarrow x=97-3^4=16.\)
\(+b=-1\text{ thì }a=b+2=1\Rightarrow\sqrt[4]{97-x}=1\Leftrightarrow x=97-x=96.\)
\(\text{Vậy }S=\left\{16;96\right\}\)
\(b\text{) ĐK: }x\ge0.\)
\(pt\Leftrightarrow\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}+\sqrt{x+4}\)
\(\Leftrightarrow x+x+9+2\sqrt{x\left(x+9\right)}=x+4+x+1+2\sqrt{\left(x+1\right)\left(x+4\right)}\)
\(\Leftrightarrow\sqrt{x^2+9x}+2=\sqrt{x^2+5x+4}\)
\(\Leftrightarrow x^2+9x+4+4\sqrt{x^2+9x}=x^2+5x+4\)
\(\Leftrightarrow x+\sqrt{x^2+9x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+\sqrt{x+9}\right)=0\)
\(\Leftrightarrow x=0\text{ (do }\sqrt{x}+\sqrt{x+9}>0\text{ }\forall x\ge0\text{)}\)
\(\text{Vậy }x=0.\)
