P=(a+2)/(2-a) – 4a^2/(a^2 -4) – (2-a)/(2+a) : (a-3)/(2a-a^2) a, rút gọn P 27/10/2021 Bởi Kylie P=(a+2)/(2-a) – 4a^2/(a^2 -4) – (2-a)/(2+a) : (a-3)/(2a-a^2) a, rút gọn P
Đáp án: $P=\dfrac{4a^2}{a-3}$ Giải thích các bước giải: $P=\left (\dfrac{a+2}{2-a}-\dfrac{4a^2}{a^2-4}-\dfrac{2-a}{2+a} \right ):\dfrac{a-3}{2a-a^2}\\=\left (-\dfrac{a+2}{a-2}-\dfrac{4a^2}{(a-2)(a+2)}-\dfrac{2-a}{2+a} \right ):\dfrac{a-3}{2a-a^2}\\=-\left (\dfrac{(a+2)^2}{(a+2)(a-2)}+\dfrac{4a^2}{(a-2)(a+2)}+\dfrac{(a-2)(2-a)}{(a-2)(2+a)} \right ).\dfrac{a(2-a)}{a-3}\\=-\left (\dfrac{a^2+4a+4}{(a+2)(a-2)}+\dfrac{4a^2}{(a-2)(a+2)}+\dfrac{2a-a^2-4+2a}{(a-2)(2+a)} \right ).\dfrac{a(2-a)}{a-3}\\=-\left (\dfrac{a^2+4a+4+4a^2+2a-a^2-4+2a}{(a+2)(a-2)} \right ).\dfrac{a(2-a)}{a-3}\\=\left (\dfrac{8a+4a^2}{(a+2)(a-2)} \right ).\dfrac{a(a-2)}{a-3}\\=\dfrac{4a(2+a)}{a+2} .\dfrac{a}{a-3}\\=4a .\dfrac{a}{a-3}\\=\dfrac{4a^2}{a-3}$ Bình luận
Đáp án:
$P=\dfrac{4a^2}{a-3}$
Giải thích các bước giải:
$P=\left (\dfrac{a+2}{2-a}-\dfrac{4a^2}{a^2-4}-\dfrac{2-a}{2+a} \right ):\dfrac{a-3}{2a-a^2}\\
=\left (-\dfrac{a+2}{a-2}-\dfrac{4a^2}{(a-2)(a+2)}-\dfrac{2-a}{2+a} \right ):\dfrac{a-3}{2a-a^2}\\
=-\left (\dfrac{(a+2)^2}{(a+2)(a-2)}+\dfrac{4a^2}{(a-2)(a+2)}+\dfrac{(a-2)(2-a)}{(a-2)(2+a)} \right ).\dfrac{a(2-a)}{a-3}\\
=-\left (\dfrac{a^2+4a+4}{(a+2)(a-2)}+\dfrac{4a^2}{(a-2)(a+2)}+\dfrac{2a-a^2-4+2a}{(a-2)(2+a)} \right ).\dfrac{a(2-a)}{a-3}\\
=-\left (\dfrac{a^2+4a+4+4a^2+2a-a^2-4+2a}{(a+2)(a-2)} \right ).\dfrac{a(2-a)}{a-3}\\
=\left (\dfrac{8a+4a^2}{(a+2)(a-2)} \right ).\dfrac{a(a-2)}{a-3}\\
=\dfrac{4a(2+a)}{a+2} .\dfrac{a}{a-3}\\
=4a .\dfrac{a}{a-3}\\
=\dfrac{4a^2}{a-3}$