a, (x^2 + 2x + 3)(x + 2)
b, (x^2 + xy + y^2)(x - y)
c, (x + y)(x^2 - xy + y^2)
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a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
a. \(\left(x+y\right)\left(x^2-xy+y^2\right)-\left(-x+y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+y^3-\left(y^3-x^3\right)\)
\(=2x^3\).
b. \(2x^3-6x^2+6x-2=2\left(x^3-3x^2+3x-1\right)=2\left(x-1\right)^3\).
b: \(C=xy\left(x^3+2\right)-y\left(xy^3+2x\right)\)
\(=x^4y+2xy-xy^4-2xy\)
\(=xy\left(x^3-y^3\right)\)
\(=xy\left(x-y\right)\left(x^2+xy+y^2\right)⋮x^2+xy+y^2\)
x+y+1=0 suy ra x+y=1
Làm câu A nhé B,C tương tự
A= x^2.(x+y-2)-(xy+y^2-2y)+(y+x-1)=0-y.(x+y-2)+1=1
Hok tốt
a)\(\dfrac{2x^2-10xy}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{2x\left(x-5y\right)}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x-5y}{y}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x\left(x-5y\right)+x\left(5y-x\right)+y\left(x+2y\right)}{xy}\)
\(=\dfrac{x^2-5xy+5xy-x^2+xy+2y^2}{xy}\)
\(=\dfrac{y\left(x+2y\right)}{xy}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{x^2+3}{2-2x^2}\)
\(=\dfrac{x+1}{2x-2}-\dfrac{x^2+3}{2x^2-2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x^2-1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)-\left(x^2+3\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
e) \(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{\left(2x^2-xy\right)-\left(xy+y^2\right)+\left(2y^2-x^2\right)}{x-y}\)
\(=\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\)
\(=\dfrac{x^2-2xy+y^2}{x-y}\)
\(=\dfrac{\left(x-y\right)^2}{x-y}\)
\(=x-y\)
a, \(\left(x^2+2x+3\right)\left(x+2\right)=x^3+2x^2+3x+2x^2+4x+6=x^3+4x^2+7x+6\)
b, \(\left(x^2+xy+y^2\right)\left(x-y\right)=x^3-y^3\)
c, \(\left(x+y\right)\left(x^2-xy+y^2\right)=x^3+y^3\)
a: Ta có: \(\left(x^2+2x+3\right)\left(x+2\right)\)
\(=x^3+2x^2+2x^2+4x+3x+6\)
\(=x^3+4x^2+7x+6\)
b: Ta có: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
c: Ta có: \(\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)
\(=x^3+y^3\)