ko ghi lai de
a/
\(=\dfrac{4}{3}a^2bc-2a^2c-\dfrac{9}{2}a^2b+2a^2c+a^2b-a^2bc\\ =\left(\dfrac{4}{3}a^2bc-a^2bc\right)+\left(-2a^2c+2a^cc\right)+\left(-\dfrac{9}{2}a^2b+2a^2b\right)\\ =\dfrac{1}{3}a^2bc-\dfrac{5}{2}a^2b\)
b/
\(=2ab-2bc^2+ab+\dfrac{1}{2}bc^2-4cb^2+2cb^2\\ =\left(2ab+ab\right)+\left(-2bc^2+\dfrac{1}{2}bc^2\right)+\left(-4cb^2+cb^2\right)\\ =3ab-\dfrac{3}{2}bc^2-3cb^2\)
c/ chưa làm dạng này nên làm như suy nghĩ thôi
=\(\dfrac{x}{3}+\dfrac{x}{6}+\dfrac{3x}{2}+\left(-\dfrac{4}{3}mn^2+0,2mn^2-1\dfrac{1}{3}mn^2\right)\)
\(=\dfrac{x}{3}+\dfrac{x}{6}+\dfrac{3x}{2}+\left(-\dfrac{4}{3}mn^2+\dfrac{2}{5}mn^2-\dfrac{4}{3}mn^2\right)\)
=\(\dfrac{x}{3}+\dfrac{x}{6}+\dfrac{3x}{2}+-\dfrac{34}{15}mn^2\)
d/
\(=\dfrac{6}{5}a^2bc-3a^2bc-\dfrac{1}{2}a^2bc+a-5-2a+7+a+2\\ =\left(\dfrac{6}{5}a^2bc-3a^2bc-\dfrac{1}{2}a^2bc\right)+\left(a-2a+a\right)+\left(-5+7+2\right)\\ =-\dfrac{23}{10}a^2bc+4\)
e/
\(=\dfrac{4}{9}a^2c^2.c^2-\dfrac{2}{5}a\left(c^2\right)^2+\dfrac{2}{3}ac^4-\dfrac{1}{4}ac^4\\ =\dfrac{4}{9}a^2c^4-\dfrac{2}{5}ac^4+\dfrac{2}{3}ac^4+\dfrac{1}{4}ac^4\\ =\dfrac{4}{9}a^2c^4+\left(-\dfrac{2}{5}ac^4+\dfrac{2}{3}ac^4+\dfrac{1}{4}ac^4\right)\\ =\dfrac{4}{9}a^2c^4+\dfrac{31}{60}ac^4\)
