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\(A=\frac{3x^2+5xy-2y^2}{3x^2-7xy+2y^2}=\frac{6xy-2y^2+3x^2-xy}{2y^2-6xy-xy+3x^2}\)
\(=\frac{2y\left(3x-y\right)+x\left(3x-y\right)}{2y\left(y-3x\right)-x\left(y-3x\right)}\)
\(=\frac{\left(3x-y\right)\left(2y+x\right)}{\left(y-3x\right)\left(2y-x\right)}=\frac{-1\left(3x-y\right)\left(2y+x\right)}{\left(y-3x\right)\left(-1\right)\left(2y-x\right)}\)
\(=\frac{\left(-3x+y\right)\left(2y+x\right)}{\left(y-3x\right)\left(-2y+x\right)}=\frac{\left(y-3x\right)\left(2y+x\right)}{\left(y-3x\right)\left(x-2y\right)}=\frac{2y+x}{x-2y}\)
\(A=\frac{y^3-x^3}{x^3-3x^2y+3xy^2-y^3}\)
\(A=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x-y\right)^3}\)
\(A=\frac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x-y\right)^2}\)
\(A=\frac{-x^2-xy-y^2}{x^2-2xy+y^2}\)
\(a,\dfrac{6x^2y^2}{8xy^5}=\dfrac{2x}{4y^3}\)
\(b,\dfrac{x^2-xy}{5xy-5y^2}=\dfrac{x\left(x-y\right)}{5y\left(x-y\right)}\)
\(c,\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x^2+1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\)
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) \(\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x+1\right)\left(x-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\)
Ta có : 3x2 - 7xy + 4y2 = 0
=> 3x2 - 3xy - 4xy + 4y2 = 0
=> 3x( x - y) - 4y( x - y) = 0
=> ( x - y)( 3x - 4y) = 0
=> x = y ; 3x = 4y
Thay : x = y ; 3x = 4y vào phân thức trên ta có:
\(A=\dfrac{4y+2x}{5y-7x}+\dfrac{3x-2y}{10y-4x}\)
\(A=\dfrac{3x+2x}{5x-7x}+\dfrac{4y-2y}{10x-4x}\)
\(A=\dfrac{5x}{-2x}+\dfrac{2y}{6x}=\dfrac{5}{-2}+\dfrac{1}{3}=\dfrac{-13}{6}\)
\(A=\dfrac{3x^2+5xy-2y^2}{3x^2-7xy+2y^2}\)
\(=\dfrac{3x^2-xy+6xy-2y^2}{3x^2-xy-6xy+2y^2}\)
\(=\dfrac{x\left(3x-y\right)+2y\left(3x-y\right)}{x\left(3x-y\right)-2y\left(3x-y\right)}\)
\(=\dfrac{\left(x+2y\right)\left(3x-y\right)}{\left(x-2y\right)\left(3x-y\right)}\)
\(=\dfrac{x+2y}{x-2y}\)