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a)\(5x^2-4\left(x^2-2x+1\right)-5=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
b) \(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)=\left(3x-2y+2\right)\left(3x+2y\right)\)
a: \(5x^2-4\left(x^2-2x+1\right)-5\)
\(=5x^2-4x^2+8x-4-5\)
\(=x^2+8x-9\)
\(=\left(x+9\right)\left(x-1\right)\)
b: \(9x^2+6x-4y^2+4y\)
\(=\left(3x+2y\right)\left(3x-2y\right)+2\left(3x+2y\right)\)
\(=\left(3x+2y\right)\left(3x-2y+2\right)\)
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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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\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)
\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)
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\(=x\left[x^2\left(x-y\right)^2-36y^2\right]\\ =x\left[x\left(x-y\right)-6y\right]\left[x\left(x-y\right)+6y\right]\\ =x\left(x^2-xy-6y\right)\left(x^2-xy+6y\right)\)
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\(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1-x-y\right)\left(xy+1+x+y\right)=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy-x-y+1\right)\left(xy+1+x+y\right)\)
\(=\left(y-1\right)\left(x-1\right)\left(y+1\right)\left(x+1\right)\)
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\(\left(x+y+z\right)^2+\left(x+y-z\right)^2-4z^2=\left(x+y+z\right)^2+\left(x+y-z-2z\right)\left(x+y-z+2z\right)=\left(x+y+z\right)^2+\left(x+y-3z\right)\left(x+y+z\right)=\left(x+y+z\right)\left(x+y+z+x+y-3z\right)=\left(x+y+z\right)\left(2x+2y-2z\right)=2\left(x+y+z\right)\left(x+y-z\right)\)
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)
\(\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)
\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)
\(=\left(x^2+5xy\right)^2+10y^2\left(x^2+5xy\right)+24y^4+y^4\)
\(=\left(x^2+5xy+5y^2\right)^2\)
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\(2x^2+x-6\)
\(=2x^2-3x+4x-6\)
\(=x\left(2x-3\right)+2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(x+2\right)\)
\(5x^2-x+y-5y^2\)
\(=\left(5x^2-5y^2\right)-\left(x-y\right)\)
\(=5\left(x^2-y^2\right)-\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left[5\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(5x+5y-1\right)\)