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Question Number 188079 by 073 last updated on 25/Feb/23
Commented by 073 last updated on 25/Feb/23
f(xx2+x+1)=x2x4+x2+xf(x)=?pleasesolution
Commented by mnjuly1970 last updated on 25/Feb/23
xx2+x+1=t⇒x2+x+1x=1tx+1x=1−tt⇒∧2x2+1x2+2=1−2t+t2t2x2+1x2+1=1−2tt2x4+x2+1x2=1−2tt2x21+x2+x4=t21−2t∴f(t)=t21−2t
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