\(\left(\dfrac{1}{a^2+a}-\dfrac{1}{a+1}\right):\dfrac{1-a}{a^2+2a+1}=\left(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{a+1}\right);\dfrac{1-a}{\left(a+1\right)^2}=\left(\dfrac{1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\right):\dfrac{1-a}{\left(a+1\right)^2}=\left(\dfrac{1-a}{a\left(a+1\right)}\right).\dfrac{\left(a+1\right)^2}{1-a}=\dfrac{a+1}{a}\)
Câu 1 :
a, A = \(=3\sqrt{8}-\sqrt{8}=2\sqrt{8}\)
b, đk a khác 0 ; a khác -1 ; 1
\(B=\left(\dfrac{1-a}{a^2+a}\right):\dfrac{1-a}{a^2+2a+1}=\dfrac{a+1}{a}\)
Câu 2 :
(d) đi qua A(2;7) <=> \(2m+n=7\left(1\right)\)
(d) đi qua B(1;3) <=> \(m+n=3\left(2\right)\)
Từ (1) ; (2) => \(\left\{{}\begin{matrix}2m+n=7\\m+n=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=4\\n=-1\end{matrix}\right.\)
Vậy ptđt (d) có dạng 4x - 1 = y
1b)Với a khác 0 a khác +-1
B=\(\left(\dfrac{1}{a^2+a}-\dfrac{1}{a+1}\right):\dfrac{1-a}{a^2+2a+1}\)
=\(\dfrac{1-a}{a^2+a}:\dfrac{1-a}{\left(a+1\right)^2}\)
=\(\dfrac{1-a}{a\left(a+1\right)}.\dfrac{\left(a+1\right)^2}{1-a}\)
=\(\dfrac{a+1}{a}\)
