Giải bài 14 trang 39

Đề bài

Thực hiện phép tính:

a) $\frac{24y^5}{7x^2}.\left(-\frac{49x}{12y^3}\right)$

b) $-\frac{36y^3}{15x^4}.\left(-\frac{45x^2}{9y^3}\right)$

c) $\frac{x^2-y^2}{x^2}.\frac{x^4}{{\left(x+y\right)}^2}$

d) $\frac{x+3}{x^2-1}.\frac{1-3x+3x^2-x^3}{9x+27}$

Lời giải chi tiết

a) $\frac{24y^5}{7x^2}.\left(-\frac{49x}{12y^3}\right)=\frac{12y^3.2y^2}{7x^2}.\left(-\frac{7x.7}{12y^3}\right)=-\frac{14y^2}{x}$

b) $-\frac{36y^3}{15x^4}.\left(-\frac{45x^2}{9y^3}\right)=-\frac{9y^3.4}{15x^2.x^2}.\left(-\frac{15x^2.3}{9y^3}\right)=\frac{12}{x^2}$

c) $\frac{x^2-y^2}{x^2}.\frac{x^4}{{\left(x+y\right)}^2}=\frac{\left(x+y\right)\left(x-y\right)}{x^2}.\frac{x^2.x^2}{{\left(x+y\right)}^2}=\frac{x^2\left(x-y\right)}{x+y}$

d) $\frac{x+3}{x^2-1}.\frac{1-3x+3x^2-x^3}{9x+27}=\frac{x+3}{\left(x+1\right)\left(x-1\right)}.\frac{-{\left(x-1\right)}^3}{9\left(x+3\right)}=-\frac{{\left(x-1\right)}^2}{9\left(x+1\right)}$