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\(\left(x+y+z\right)^2+\left(x+y-z\right)^2-4z^2=\left(x+y+z\right)^2+\left(x+y-z-2z\right)\left(x+y-z+2z\right)=\left(x+y+z\right)^2+\left(x+y-3z\right)\left(x+y+z\right)=\left(x+y+z\right)\left(x+y+z+x+y-3z\right)=\left(x+y+z\right)\left(2x+2y-2z\right)=2\left(x+y+z\right)\left(x+y-z\right)\)
Ta có:
(x + y + z)2 + (x + y – z)2 – 4z2
\(=\left(x+y-z\right)^2+\left(x+y-z\right)\left(x+y+3z\right)\)
\(=\left(x+y-z\right)\left(x+y+3z+x+y-z\right)\)
\(=2\left(x+y-z\right)\left(x+y+z\right)\)![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x\left[x^2\left(x-y\right)^2-36y^2\right]\\ =x\left[x\left(x-y\right)-6y\right]\left[x\left(x-y\right)+6y\right]\\ =x\left(x^2-xy-6y\right)\left(x^2-xy+6y\right)\)
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\(A=-x-z\left(x-y\right)+y=-x-xz+zy+y=-x\left(1+z\right)+y\left(1+z\right)=\left(1+z\right)\left(y-x\right)\)
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\(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1-x-y\right)\left(xy+1+x+y\right)=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy-x-y+1\right)\left(xy+1+x+y\right)\)
\(=\left(y-1\right)\left(x-1\right)\left(y+1\right)\left(x+1\right)\)
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\(\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)
\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)
\(=\left(x^2+5xy\right)^2+10y^2\left(x^2+5xy\right)+24y^4+y^4\)
\(=\left(x^2+5xy+5y^2\right)^2\)
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a) x^2-9x+20
= x^2+4x+5x+20
=(x^2+4x)+(5x+20)
=x(x+4)+5(x+4)
=(x+4)(x+5)
\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)
\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)