bài 7 :cho các đơn thức A = -x ² y ³,B=-2/9xy ²,C=-3y+2x tính :a)a.c+b b)b.c-a c)a.b.c d)a/b.c 05/09/2021 Bởi Peyton bài 7 :cho các đơn thức A = -x ² y ³,B=-2/9xy ²,C=-3y+2x tính :a)a.c+b b)b.c-a c)a.b.c d)a/b.c
$a) A.C+B=(-x^{2}y^{3})(-3y+2x)+\left ( -\frac{2}{9}xy \right )\\=-x^{2}y^{3}(-3y+2x)-\frac{2}{9}xy\\=3x^{2}y^{4}-2x^{3}y^{3}-\frac{2}{9}xy\\b) B.C-A=\left ( -\frac{2}{9}xy \right )(-3y+2x)-(-x^{2}y^{3})\\= -\frac{2}{9}xy(-3y+2x)+x^{2}y^{3}\\=\frac{2}{3}xy^{2}-\frac{4}{9}x^{2}y+x^{2}y^{3}\\c) \frac{A}{B.C}=\frac{-x^{2}y^{3}}{\left ( -\frac{2}{9}xy \right )(-3y+2x)}\\=\frac{xy^{2}}{\frac{2}{9}(-3y+2x)}\\=\frac{xy^{2}}{-\frac{2}{3}y+\frac{4}{9}x}\\=\frac{xy^{2}}{\frac{-6y+4x}{9}}\\=\frac{9xy^{2}}{-6y+4x}$ Bình luận
$a) A.C+B=(-x^{2}y^{3})(-3y+2x)+\left ( -\frac{2}{9}xy \right )\\=-x^{2}y^{3}(-3y+2x)-\frac{2}{9}xy\\=3x^{2}y^{4}-2x^{3}y^{3}-\frac{2}{9}xy\\b) B.C-A=\left ( -\frac{2}{9}xy \right )(-3y+2x)-(-x^{2}y^{3})\\= -\frac{2}{9}xy(-3y+2x)+x^{2}y^{3}\\=\frac{2}{3}xy^{2}-\frac{4}{9}x^{2}y+x^{2}y^{3}\\c) \frac{A}{B.C}=\frac{-x^{2}y^{3}}{\left ( -\frac{2}{9}xy \right )(-3y+2x)}\\=\frac{xy^{2}}{\frac{2}{9}(-3y+2x)}\\=\frac{xy^{2}}{-\frac{2}{3}y+\frac{4}{9}x}\\=\frac{xy^{2}}{\frac{-6y+4x}{9}}\\=\frac{9xy^{2}}{-6y+4x}$