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\(a,\dfrac{6x^2y^2}{8xy^5}=\dfrac{2x}{4y^3}\)
\(b,\dfrac{x^2-xy}{5xy-5y^2}=\dfrac{x\left(x-y\right)}{5y\left(x-y\right)}\)
\(c,\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x^2+1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(\dfrac{x^2-xy}{5xy-5y^2}=\dfrac{x\left(x-y\right)}{5y\left(x-y\right)}=\dfrac{x}{5y}\)
c) \(\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x+1\right)\left(x-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
a) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)
b)\(=\frac{3x\left(x+y\right)}{y}\)
c) \(\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=-\frac{7}{3}.\)
b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)
c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\frac{x-z}{2}\)
h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}=-\frac{3x\left(x-1\right)}{2\left(x-1\right)}=\frac{-3x}{2}\)
j) \(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)
Câu b) bạn xem lại nhé.
Học tốt ^3^
a) \(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)=\(\frac{\left(x-1\right)\left(x-y\right)}{\left(x-1\right)\left(x+y\right)}\)=\(\frac{x-y}{x+y}\)
b) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
c) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
a)\(\frac{3xy+6}{6xy+12}=\frac{1}{2}\Leftrightarrow\left(3xy+6\right)\cdot2=\left(6xy+12\right)\cdot1\)
\(\Leftrightarrow6xy+12=6xy+12\)
Vậy.......
b)\(\frac{x^2-xy}{5y^2-5xy}=\frac{x}{5y}\Leftrightarrow\left(x^2-xy\right)\cdot5y=\left(5y^2-5xy\right)\cdot x\)
\(\Leftrightarrow5x^2y-5xy^2=5xy^2-5x^2y\)
Vậy.....
a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)
\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)
\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)
b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)
\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)
\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)
\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)
\(\Leftrightarrow\frac{1}{3x^2-y^2}\)
=\(\frac{30x^2y^3}{8x^2y^2}\)
Ta có :
\(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)
\(\frac{x^2-xy}{5xy-5y^2}=\frac{x\left(x-y\right)}{5y\left(x-y\right)}=\frac{x}{5y}\)
Hok tốt !