b) (2x + 1)(y + 1) = 6
c) xy + x = 5
d) xy + x + y = 2
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a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
b: \(C=xy\left(x^3+2\right)-y\left(xy^3+2x\right)\)
\(=x^4y+2xy-xy^4-2xy\)
\(=xy\left(x^3-y^3\right)\)
\(=xy\left(x-y\right)\left(x^2+xy+y^2\right)⋮x^2+xy+y^2\)
a: \(=\dfrac{x^2-1-3x^2+3+2x^2+7}{2x-y}=\dfrac{9}{2x-y}\)
b: \(=\dfrac{x+y+x-y+2x-3y}{1-xy}=\dfrac{4x-3y}{1-xy}\)
a, \(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}=\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(y-x\right)}\)
\(=\frac{x^2}{xy\left(x-y\right)}-\frac{2xy-y^2}{xy\left(x-y\right)}=\frac{\left(x-y\right)^2}{xy\left(x-y\right)}=\frac{x-y}{xy}\)
b, \(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x^2}{x^2-1}=\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1+x+1+2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{2x+2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2x}{x-1}\)
c: =>x(y+1)=5
\(\Leftrightarrow\left(x,y+1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(1;4\right);\left(5;0\right);\left(-1;-6\right);\left(-5;-2\right)\right\}\)