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\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)
\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)
\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)
\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)
\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-x-y+1\right)\)
\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)
\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)
\(x^2+2xy+x+2y\)
\(=x\left(x+1\right)+2y\left(x+1\right)\)
\(=\left(x+1\right)\left(2y+x\right)\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
a)x2+2xy+x+2y
=(2xy+x2)+(2y+x)
=x(2y+x)+(2y+x)
=(x+1)(2y+x)
b)7x2-7xy-5x+5y
=(5y-7xy)+(7x2-5x)
=y(5-7x)-x(5-7x)
=(5-7x)(y-x)
c)x2-6x+9-9y2
=(x2+3xy-3x)-(3xy+9y2-9y)-(3x+9y-9)
=x(x+3y-3)-3y(x+3y-3)-3(x+3y-3)
=(x-3y-3)(x+3y-3)
d)x3-3x2+3x-1+2(x2-x)
Ta thấy x=1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x-1
=(x-1)(x2+1)
e) (x+y)(y+z)(z+x)+xyz
đề sai
f)x(y2-z2)+y(z2-x2)
=(xy2+yz2)+(x2y+xz2)
=y(xy+z2)-x(xy+z2)
=(y-x)(xy+z2)
\(x;y;z\ne0\). Giả thiết của đề bài:
\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{z+x}\Leftrightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{x+z}{xz}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{x}+\frac{1}{z}\Leftrightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}.\)
=> x = y = z
Do đó, M = 1.
a) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
x2−x−y2−y=(x2−y2)−(x+y)=(x−y)(x+y)−(x+y)=(x+y)(x−y−1)
x^2-2xy+y^2-z^2
= (x-y)^2 - z^2
= (x-y-z)(x-y+z)
5x-5y+5x-ay
xy(x+y)+yz(y+z)+xz(x+z)+2xyz