Đáp án:
$a)A=\dfrac{62}{9}\\ b)B=-3\\ c)C=\dfrac{20}{9}\\ d)D=-8$
Giải thích các bước giải:
$a)A=3x^3-6x^2+2|x|+7\\ x=-\dfrac{1}{3} \, A=3\left(-\dfrac{1}{3} \right)^3-6\left(-\dfrac{1}{3} \right)^2+2\left|-\dfrac{1}{3}\right|+7\\ =-3.\dfrac{1}{3^3}-6.\dfrac{1}{3^2} +2.\dfrac{1}{3}+7\\ =-\dfrac{1}{9}-\dfrac{2}{3} +\dfrac{2}{3}+7\\ =-\dfrac{1}{9}+7\\ =-\dfrac{1}{9}+\dfrac{63}{9}\\ =\dfrac{62}{9}\\ b)B=4\left|x\right|-2\left|y\right|\\ x=\dfrac{1}{4},y=-2, B=4\left|\dfrac{1}{4}\right|-2\left|-2\right|\\ =4.\dfrac{1}{4}-2.2\\ =1-4\\ =-3\\ c)C=6x^3-3x^2+2\left|x\right|+4\\ x=-\dfrac{2}{3}, C=6.\left(-\dfrac{2}{3} \right)^3-3.\left(-\dfrac{2}{3} \right)^2+2\left|-\dfrac{2}{3}\right|+4\\ =-6.\dfrac{2^3}{3^3}-3.\dfrac{2^2}{3^2}+2.\dfrac{2}{3}+4\\ =-\dfrac{2^4}{3^2}-\dfrac{4}{3}+\dfrac{4}{3}+4\\ =-\dfrac{16}{9}+4\\ =-\dfrac{16}{9}+\dfrac{36}{9}\\ =\dfrac{20}{9}\\ d)D=2\left|x\right|-3\left|y\right|\\ x=\dfrac{1}{2},y=-3, D=2\left|\dfrac{1}{2}\right|-3\left|-3\right|\\ =2.\dfrac{1}{2}-3.3\\ =1-9\\ =-8$
