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If $\alpha, \beta$ be the roots of $x^2-a(x-1)+b=0$, then the value of $\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}$
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$$
\begin{aligned}
& \text { Hints }: x^2-a x=a+3 \quad \alpha \beta=a+b \\
& \alpha+\beta=a \\
& \alpha^2-a \alpha=-(a+b) \\
& \beta^2-a \alpha=-(a+b) \\
&-\frac{1}{a+b}-\frac{1}{a+b}+\frac{2}{a+b}=0
\end{aligned}
$$
\begin{aligned}
& \text { Hints }: x^2-a x=a+3 \quad \alpha \beta=a+b \\
& \alpha+\beta=a \\
& \alpha^2-a \alpha=-(a+b) \\
& \beta^2-a \alpha=-(a+b) \\
&-\frac{1}{a+b}-\frac{1}{a+b}+\frac{2}{a+b}=0
\end{aligned}
$$
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