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\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
\(=\left[\dfrac{\left(3x+y\right)\left(x+3y\right)+\left(3x-y\right)\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\right].\dfrac{\left(x-3y\right)\left(x+3y\right)}{x^2+y^2}\)
\(=\dfrac{\left(3x+y\right)\left(x+3y\right)+\left(3x-y\right)\left(x-3y\right)}{x.\left(x^2+y^2\right)}\)
\(=\dfrac{3x^2+3xy+xy+3y^2+3x^2-3xy-xy+3y^2}{x\left(x^2+y^2\right)}\)
\(=\dfrac{6x^2+6y^2}{x\left(x^2+y^2\right)}=\dfrac{6\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{6}{x}\)
a: \(=\dfrac{x^2+xy-x^2-y^2}{x+y}\cdot\dfrac{x-y+2y}{y\left(x-y\right)}\)
\(=\dfrac{y\left(x-y\right)}{x+y}\cdot\dfrac{x+y}{y\left(x-y\right)}=1\)
b: \(\left(\dfrac{2}{x^2-1}+\dfrac{x^2-3}{3x^2-1}\right):\left[\dfrac{1}{x}-\dfrac{2x\left(x^2-3\right)}{\left(x^2-1\right)\left(3x^2-1\right)}\right]\)
\(=\dfrac{6x^2-2+x^4-4x^2+3}{\left(x^2-1\right)\left(3x^2-1\right)}:\dfrac{\left(x^2-1\right)\left(3x^2-3\right)-2x^2\left(x^2-3\right)}{x\left(x^2-1\right)\left(3x^2-1\right)}\)
\(=\dfrac{x^4+2x^2+1}{\left(x^2-1\right)\left(3x^2-1\right)}\cdot\dfrac{x\left(x^2-1\right)\left(3x^2-1\right)}{3x^4-6x^2+3-2x^4+6x^2}\)
\(=\dfrac{x\left(x^2+1\right)^2}{x^4+3}\)
a: \(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{3}{x-3}\cdot\dfrac{-\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{-3}{x-3}\)
b: \(=\dfrac{x+1}{x+2}:\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)^2}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x^2+2xy+y^2}{2xy}\cdot\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{2\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)^2}{x^2+y^2}\cdot\dfrac{1}{2}\)
\(=\dfrac{\left(x+y\right)}{x-y}\)
\(a.\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{-2\left(y-x\right)}{3\left(y-x\right)}=\dfrac{-2}{3}\)
\(b.\dfrac{x-2}{-x}=\dfrac{2-x}{x}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{8-x^3}{x\left(x^2+2x+4\right)}\)
\(\dfrac{3x}{x+y}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{-3x\left(x-y\right)}{\left(x+y\right)\left(y-x\right)}=\dfrac{-3x\left(x-y\right)}{y^2-x^2}\)
c: \(\dfrac{-3x\left(x-y\right)}{y^2-x^2}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{3x}{x+y}\)
a: \(\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{2\left(x-y\right)}{-3\left(x-y\right)}=\dfrac{-2}{3}\)
b: \(\dfrac{8-x^3}{x\left(x^2+2x+4\right)}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{2-x}{x}\)
a)2(x-y)/(-3)(x-y)=-2/3
b)8-x^3=(2-x)(x^2+2x+4) => Vế phải =(2-x)/x=(x-2)/-x
c)y^2-x^2=(y+x)(y-x) bạn đổi dấu rồi rút gọn là được,cũng tương tự như trên ý