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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) \(\frac{x^2+5x+6}{x^2+7x+12}\)=\(\frac{x^2+2x+3x+6}{x^2+3x+4x+12}\)=\(\frac{x\left(x+2\right)+3\left(x+2\right)}{x\left(x+3\right)+4\left(x+3\right)}\)=\(\frac{\left(x+3\right)\left(x+2\right)}{\left(x+4\right)\left(x+3\right)}\)
b) \(\frac{7x^2+14x+7}{3x^2+3x}\)=\(\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)\left(x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)}{3x}\)
a)\(\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}\)=\(\frac{x+1}{5x^2}\)
b)\(\frac{10y}{15\left(x+y\right)^2}\)
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
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.....
Không ai trả lời buồn quá .
\(\frac{3x^2-12x+12}{x^4-8x}\)
\(=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+2^2\right)}\)
\(=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
bn hãy xem cách tui làm phía duoi la làm dc, cứ xem có j chung thi rút nó ra
a) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)