Tách hạng tử là gì vậy? Tớ ko biết nó là gì nên lm 2 cách
a) C1: \(x^2-6x+8\)
\(=x^2-2.3x+3^2-1^2\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
C2: \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=\left(x^2-4x\right)-\left(2x-8\right)\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
b) \(x^2-4x+3\)
C1: \(=x^2-2.2x+2^2-1^2\)
\(=\left(x-2\right)^2-1^2\)
\(=\left(x-2-1\right)\left(x-2+1\right)\)
\(=\left(x-3\right)\left(x-1\right)\)
C2: \(=x^2-x-3x+3\)
\(=\left(x^2-x\right)-\left(3x-3\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\)
c) \(x^2-x-12\)
C1: \(=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}-\dfrac{7}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{7}{2}\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
C2: \(=x^2-4x+3x-12\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x+3\right)\left(x-4\right)\)
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