Câu 1:
\(a.a^2+ab+9a+9b\\ =a\left(a+b\right)+9\left(a+b\right)\\ =\left(a+9\right)\left(a+b\right)\\ b.3x\left(x+1\right)-9\left(x+1\right)=0\\ \left(3x-9\right)\left(x+1\right)=0\\ 3\left(x-3\right)\left(x+1\right)=0\\ \Rightarrow\left(x-3\right)\left(x+1\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
\(c.\dfrac{1}{ac-a^2}-\dfrac{1}{c^2-ac}\\ =\dfrac{1}{a\left(c-a\right)}-\dfrac{1}{c\left(c-a\right)}\\ =\dfrac{c}{ac\left(c-a\right)}-\dfrac{a}{ac\left(c-a\right)}\\ =\dfrac{c-a}{ac\left(c-a\right)}\\ =\dfrac{1}{ac}\)
Đúng 1
Bình luận (0)
