rút gọn:
a,\(\frac{\text{}\text{}15x^2y^4}{5x^3z}\)
b, \(\frac{x^2-4x+4}{x^2-4}\)
c,\(\frac{5x^2+10xy+5y^2}{15x+15y}\)
d,\(\frac{2x^3-2}{11x^2-22x+11}\)
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15 tháng 3 2020
1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
f)
$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$
$=\frac{2x-3x^2}{x^2-1}.\frac{x^4-1}{(2-3x)^3}=\frac{x(2-3x)(x^2-1)(x^2+1)}{(x^2-1)(2-3x)^3}$
$=\frac{x(x^2+1)}{(2-3x)^2}$
g)
$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$
$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$
h)
$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$
$=\frac{x(x+2)}{3(x-1)^2}.\frac{5(x-1)}{2(x+2)}$
$=\frac{5x}{6(x-1)}$
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
d)
$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$
$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$
$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$
$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)
$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$
$=\frac{-3(x+7)}{2x+1}$
\(a,\frac{15x^2y^4}{5x^3z}=\frac{3y^4}{x}\)
\(b,\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)
\(c,\frac{5x^2+10xy+5y^2}{15x+15y}=\frac{5\left(x^2+2xy+y^2\right)}{15\left(x+y\right)}=\frac{5\left(x+y\right)^2}{15\left(x+y\right)}=\frac{x+y}{3}\)
\(d,\frac{2x^3-2}{11x^2-22x+11}=\frac{2\left(x^3-1\right)}{11\left(x^2-2x+1\right)}=\frac{2\left(x-1\right)\left(x^2+x+1\right)}{11\left(x-1\right)^2}=\frac{2\left(x^2+x+1\right)}{11\left(x-1\right)}\)