Phân tích đa thức thành nhân tử : x2 - 2x - 24
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\(x^2-3xy+2x-6y\)
= \(x\left(x-3y\right)+2\left(x-3y\right)\)
= \(\left(x+2\right)\left(x-3y\right)\)
= ( x2 - 4y2 ) - ( 2x + 4y )
= ( x - 2y ) ( x + 2y ) - 2 ( x - 2y )
= ( x - 2y ) ( x + 2y - 2 )
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
x2 + 2x – 3
= x2 + 2x + 1 – 4
= (x + 1)2 – 22
= (x + 1 + 2)(x + 1 – 2)
= (x + 3)(x – 1)
\(x^2+2x+1-16=\left(x+1\right)^2-4^2=\left(x+1-4\right).\left(x+1+4\right)=\left(x-3\right).\left(x+5\right)\)
\(x^2+2x+1-16=\left(x^2+2x+1\right)-4^2=\left(x+1\right)^2-4^2=\left(x+1-4\right)\left(x+1+4\right)=\left(x-3\right)\left(x+5\right)\)
Cách 1: x 2 + 2xy - 15 y 2 = ( x 2 + 2xy + y 2 ) - 16 y 2
= x + y 2 - 4 y 2
= (x + y + 4y)(x + y – 4y)
= (x + 5y)(x – 3y).
Cách 2: x 2 + 2xy - 15 y 2 = x 2 + 5xy – 3xy - 15 y 2
= x(x + 5y) – 3y(x + 5y)
= (x – 3y)(x + 5y).
`x^2+2x+1-y^2+2y-1`
`=(x^2+2x+1)-(y^2-2y+1)`
`=(x+1)^2-(y-1)^2`
`=(x+1+y-1)(x+1-y+1)`
`=(x+y)(x-y+2)`
Ta có: \(x^2+2x+1-y^2+2y-1\)
\(=\left(x+1\right)^2-\left(y-1\right)^2\)
\(=\left(x+1-y+1\right)\left(x+1+y-1\right)\)
\(=\left(x-y+2\right)\left(x+y\right)\)
\(x^2-2x-24\)
\(=x^2-6x+4x-24\)
\(=x(x-6)+4(x-6)\)
\(=(x+4)(x-6)\)
\(x^2-2x-24\\ =x^2-2x+1-25\\ =\left(x-1\right)^2-5^2\\ =\left(x-1-5\right)\left(x-1+5\right)\\ =\left(x-6\right)\left(x+4\right)\)