Đáp án:
$a)\frac{36y^4+24xy^4}{120x^4y^5}\\
\frac{75x^2y^3}{120x^4y^5}\\
\frac{80x^3}{120x^4y^5}\\
c) \frac{(7x-1)(x-3)}{2x(x^2-9)}\\
\frac{2x(5-3x)}{2x(x^2-9)}\\
e) \frac{7(4y^2-x^2)}{2x(4y^2-x^2)}\\
\frac{-8x(2y+x)}{2x(4y^2-x^2)}\\
\frac{x(x-y)}{2x(4y^2-x^2)}\\
b) \frac{6(x^2-1)}{3x(x+3)(x+1)}\\
\frac{x^2-9}{3x(x+3)(x+1)}\\
d) \frac{-2(1-x^2)}{2x(1-x)^2}\\
\frac{x(x+2)}{2x(1-x)^2}\\
f) \frac{x^3y}{y(x-y)^3}\\
\frac{-x(x-y)^2}{y(x-y)^3}$
Giải thích các bước giải:
$a) MSC: 120x^4y^5\\
\frac{3+2x}{10x^4y}=\frac{(3+2x).12y^4}{120x^4y^5}=\frac{36y^4+24xy^4}{120x^4y^5}\\
\frac{5}{8x^2y^2}=\frac{15x^2y^3.5}{120x^4y^5}=\frac{75x^2y^3}{120x^4y^5}\\
\frac{2}{3xy^5}=\frac{40x^3.2}{120x^4y^5}=\frac{80x^3}{120x^4y^5}\\
c) MSC: 2x(x-3)(x+3)=2x(x^2-9)\\
\frac{7x}{2x^2+6x}=\frac{7x-1}{2x(x+3)}=\frac{(7x-1)(x-3)}{2x(x^2-9)}\\
\frac{5-3x}{x^2-9}=\frac{2x(5-3x)}{2x(x^2-9)}\\
e) MSC: 2x(2y-x)(2y+x)=2x(4y^2-x^2)\\
\frac{7}{2x}=\frac{7(4y^2-x^2)}{2x(4y^2-x^2)}\\
\frac{4}{x-2y}=\frac{-4}{2y-x}=\frac{(-4).2x.(2y+x)}{2x(4y^2-x^2)}=\frac{-8x(2y+x)}{2x(4y^2-x^2)}\\
\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2(4y^2-x^2)}=\frac{x(x-y)}{2x(4y^2-x^2)}\\
b) MSC: 3x(x+1)(x+3)\\
\frac{4x-4}{2x(x+3)}=\frac{4(x-1)}{2x(x+3)}=\frac{2(x-1)}{x(x+3)}=\frac{2.3(x-1)(x+1)}{3x(x+3)(x+1)}=\frac{6(x^2-1)}{3x(x+3)(x+1)}\\
\frac{x-3}{3x(x+1)}=\frac{(x-3)(x+3)}{3x(x+1)(x+3)}=\frac{x^2-9}{3x(x+3)(x+1)}\\
d) MSC: 2x(1-x)^2\\
\frac{x+1}{x-x^2}=\frac{x+1}{x(1-x)}=\frac{2(x+1)(1-x)}{2x(1-x)^2}=\frac{-2(1-x^2)}{2x(1-x)^2}\\
\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2(1-2x+x^2)}=\frac{x+2}{2(1-x)^2}=\frac{x(x+2)}{2x(1-x)^2}\\
f) MSC:y(x-y)^3\\
\frac{x^3}{x^2-3x^2y+3xy^2-y^3}=\frac{x^3}{(x-y)^3}=\frac{x^3y}{y(x-y)^3}\\
\frac{x}{y^2-xy}=\frac{x}{y(y-x)}=\frac{-x}{y(x-y)}=\frac{-x(x-y)^2}{y(x-y)^3}$
