1−x−−−−√4+x+15−−−−−√4=21−x4+x+154=2(ĐK: −15≤x≤1−15≤x≤1)
⇔1−x2−−−−−√1+x−−−−√4+x+15−−−−−√−4x+15−−−−−√4+2=0⇔1−x21+x4+x+15−4x+154+2=0
⇔1−x21+x−−−−√=x−1(x+15−−−−−√4+2)2⇔1−x21+x=x−1(x+154+2)2
⇔(x−1)[1(x+15−−−−−√4+2)2−(1+x)1−x−−−−√]=0⇔(x−1)[1(x+154+2)2−(1+x)1−x]=0
⇔x=1⇔x=1(Vì 1(x+15−−−−−√4+2)2−1+x1−x−−−−√1(x+154+2)2−1+x1−x