Bài 1:
\(a,ĐK:a\ne-1;a\ne-2;a\ne-3;\\ b,A=\dfrac{4a^3+12a^2+10a^2+22a+4+2a^2+26a+60}{\left(a+1\right)\left(a+2\right)\left(a+3\right)}\\ A=\dfrac{4a^3+24a^2+48a+64}{\left(a+1\right)\left(a+2\right)\left(a+3\right)}=\dfrac{4\left(a^3+6a^2+12a+16\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)}\\ A=\dfrac{4\left(a^3+4a^2+2a^2+8a+4a+16\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)}\\ A=\dfrac{4\left(a+4\right)\left(a^2+2a+4\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)}\)
Bài 2:
\(\dfrac{n^3+n^2+3}{n+1}=\dfrac{n^2\left(n+1\right)+3}{n+1}=n^2+\dfrac{3}{n+1}\in Z\\ \Leftrightarrow n+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow n\in\left\{-4;-2;0;2\right\}\left(tm\right)\)
Bài 3:
\(\text{Đẳng thức }=\dfrac{\left(x+z\right)\left(x-z\right)-\left(x+y\right)\left(x-y\right)-\left(y+z\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\\ =\dfrac{x^2-z^2-x^2+y^2-y^2+z^2}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\left(đpcm\right)\)
Bài 4:
\(VT=\dfrac{4a^2-4a+1-4a^2+4a+3}{\left(2a-1\right)\left(2a+1\right)}\cdot\dfrac{2a+1}{2a-1}=\dfrac{4}{\left(2a-1\right)^2}=\left(\dfrac{2}{2a-1}\right)^2=VP\)
