Giả thiết tương đương:
\(x.f\left(x\right)=\sqrt{x^2+1}-\left(x^2+1\right)f'\left(x\right)\)
\(\Leftrightarrow\left(x^2+1\right)f'\left(x\right)+x.f\left(x\right)=\sqrt{x^2+1}\)
\(\Leftrightarrow\sqrt{x^2+1}.f'\left(x\right)+\dfrac{x}{\sqrt{x^2+1}}f\left(x\right)=1\)
\(\Leftrightarrow\left[\sqrt{x^2+1}.f\left(x\right)\right]'=1\)
Lấy nguyên hàm 2 vế:
\(\Rightarrow\sqrt{x^2+1}.f\left(x\right)=\int1.dx=x+C\)
Thay \(x=0\Rightarrow1.f\left(0\right)=0+C\Rightarrow C=1\)
\(\Rightarrow\sqrt{x^2+1}.f\left(x\right)=x+1\)
\(\Rightarrow f\left(x\right)=\dfrac{x+1}{\sqrt{x^2+1}}\)
\(\Rightarrow f\left(1\right)=\sqrt{2}\)
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