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\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)
ĐKXĐ: \(x\ne y\)
a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)
b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)
\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)
SỬa đề: x^3-xy^2
\(A=\left(\dfrac{x\left(x-y\right)}{y\left(x+y\right)}+\dfrac{x^2-y}{x\left(x+y\right)}\right):\left(\dfrac{y^2}{x\left(x^2-y^2\right)}+\dfrac{1}{x-y}\right)\)
\(=\left(\dfrac{x^2\left(x-y\right)+y\left(x^2-y\right)}{xy\left(x+y\right)}\right):\left(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}+\dfrac{x\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^3-x^2y+x^2y-y^3}{xy\left(x+y\right)}:\dfrac{y^2+x^2+xy}{x\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\cdot\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+xy+y^2}=\dfrac{\left(x-y\right)^2}{y}\)
Để A>0 thì y>0
a: F=9/25x^2y^4*20/27x^3y=4/15x^5y^5
Bậc: 10
b: y=-x/3 và x+y=2
=>x+y=2 và -1/3x-y=0
=>x=3 và y=-1
Khi x=3 và y=-1 thì F=4/15*(-3)^5=-324/5
\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)
\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)
