Google
Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $R\backslash \left\{ -2;0 \right\}$ thỏa mãn $x.\left( x+2 \right).f'\left( x \right)+2f\left( x \right)={{x}^{2}}+2x$ và $f\left( 1 \right)=-6\ln 3$. Biết $f\left( 3 \right)=a+b.\ln 5$ $\left( a,b\in \mathbb{Q} \right)$.
Giá trị của $a-b$ bằng
A. $20$.
B. $10$.
C. $\dfrac{10}{3}$.
D. $\dfrac{20}{3}$.
Ta có $x.\left( x+2 \right).f'\left( x \right)+2f\left( x \right)={{x}^{2}}+2x\Leftrightarrow f'\left( x \right).\dfrac{x}{x+2}+\dfrac{2}{{{\left( x+2 \right)}^{2}}}.f\left( x \right)=\dfrac{x}{x+2}$
$\Leftrightarrow {{\left( f\left( x \right).\dfrac{x}{x+2} \right)}^{'}}=\dfrac{x}{x+2}\Rightarrow f\left( x \right).\dfrac{x}{x+2}=\int{\dfrac{x}{x+2}dx}\Rightarrow f\left( x \right).\dfrac{x}{x+2}=x-2\ln \left( x+2 \right)+C$
Do $f\left( 1 \right)=-6\ln 3\Rightarrow C=-1$ $\Rightarrow f\left( x \right).\dfrac{x}{x+2}=x-2\ln \left( x+2 \right)-1\Rightarrow f\left( 3 \right)=\dfrac{10}{3}-\dfrac{10}{3}\ln 5$.
$\Rightarrow a=\dfrac{10}{5};b=\dfrac{-10}{5}\Rightarrow a-b=\dfrac{20}{5}$.
Giá trị của $a-b$ bằng
A. $20$.
B. $10$.
C. $\dfrac{10}{3}$.
D. $\dfrac{20}{3}$.
Ta có $x.\left( x+2 \right).f'\left( x \right)+2f\left( x \right)={{x}^{2}}+2x\Leftrightarrow f'\left( x \right).\dfrac{x}{x+2}+\dfrac{2}{{{\left( x+2 \right)}^{2}}}.f\left( x \right)=\dfrac{x}{x+2}$
$\Leftrightarrow {{\left( f\left( x \right).\dfrac{x}{x+2} \right)}^{'}}=\dfrac{x}{x+2}\Rightarrow f\left( x \right).\dfrac{x}{x+2}=\int{\dfrac{x}{x+2}dx}\Rightarrow f\left( x \right).\dfrac{x}{x+2}=x-2\ln \left( x+2 \right)+C$
Do $f\left( 1 \right)=-6\ln 3\Rightarrow C=-1$ $\Rightarrow f\left( x \right).\dfrac{x}{x+2}=x-2\ln \left( x+2 \right)-1\Rightarrow f\left( 3 \right)=\dfrac{10}{3}-\dfrac{10}{3}\ln 5$.
$\Rightarrow a=\dfrac{10}{5};b=\dfrac{-10}{5}\Rightarrow a-b=\dfrac{20}{5}$.
Đáp án D.