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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) \(\dfrac{36\left(x-2\right)^3}{32-16x}=\dfrac{36\left(x-2\right)^3}{16\left(2-x\right)}=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}\)\(=\dfrac{36\left(x-2\right)^3:4\left(x-2\right)}{-16\left(x-2\right):4\left(x-2\right)}\)\(=\dfrac{9\left(x-2\right)^2}{-4}\)
b) \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{x\left(x-y\right)}{-5y\left(x-y\right)}\)\(=\dfrac{x}{-5y}\)
\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=0-11x+24\)
\(=-11x+24\)
\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)
\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)
\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)
\(=0+0+5\)
\(=5\)
a) \(=12y^2+3y+28-12y^2=3y+28\)
b) \(=x^2-4x+4-3x^2+8x+3=-2x^2+4x+7\)
\(a.\)
\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)
\(b.\)
\(\dfrac{4x^2-4xy+y^2}{-\left(4x^2-y^2\right)}\\ =-\dfrac{\left(2x-y\right)^2}{\left(2x-y\right)\left(2x+y\right)}\\ =\dfrac{-\left(2x-y\right)}{2x+y}\\ =\dfrac{y-2x}{y+2x}\)
a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)
\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)
\(=\dfrac{4x+1}{4x-1}\)
b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)
\(=\dfrac{\left(2x-y\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{\left(y-2x\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{y-2x}{y+2x}\)
\(\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}-\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9}{x\left(x-3\right)}\right)\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ =\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6}{x}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-3}{x-1}\)