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\(A=9x^2+3yx+6x+y+1\)
\(\Rightarrow A=\left(9x^2+3x\right)+\left(3xy+y\right)+\left(3x+1\right)\)
\(\Rightarrow A=3x\left(3x+1\right)+y\left(3x+1\right)+\left(3x+1\right)\)
\(\Rightarrow A=\left(3x+y+1\right)\left(3x+1\right)\)
TL:
\(A=9x^2-y^2+6x+1\)
\(=\left(3x-1\right)^2-y^2\)
\(=\left(3x-1+y\right)\left(3x-1-y\right)\)
\(9x^2-6x+1\)
\(=\left(3x\right)^2-2.3x.1+1^2\)
\(=\left(3x-1\right)^2\)
5x^2+10xy+5y^2
=5.(x2+2xy+y2)
=5.(x+y)2
x^3-6x^2+9x
=x.(x2-6x+9)
=x.(x-3)2
xy+y^2-x-y
=y.(x+y)-(x+y)
=(x+y)(y-1)
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
\(9x^2-6x^2-3\)
\(=3x^2-3\)
\(=3.\left(x^2-1\right)\)
\(=3.\left(x-1\right).\left(x+1\right)\)
\(9x^2-6x^2-3\)
\(=3x^2-3\)
\(=3.\left(x^2-1\right)\)
\(=3.\left(x-1\right).\left(x+1\right)\)
Nguồn: kudo shinichi
\(-9x^2+6x+y^2-1\)
\(=-\left(9x^2-6x+1-y^2\right)\)
\(=-\left(3x-1-y\right)\left(3x-1+y\right)\)