a) \(x^2-3x+2\)
\(\Leftrightarrow x^2-2x-x+2\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\)
b) \(x^2-6x+8\)
\(\Leftrightarrow x^2-4x-2x+8\)
\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x-2\right)\)
c) \(3x^2+9x-30\)
\(\Leftrightarrow3\left(x^2+3x-10\right)\)
\(\Leftrightarrow3\left[\left(x^2+2\cdot\frac{3x}{2}+\frac{9}{4}\right)-\frac{49}{4}\right]\)
\(\Leftrightarrow3\left[\left(x+\frac{3}{2}\right)^2-\left(\frac{7}{2}\right)^2\right]\)
\(\Leftrightarrow3\left(x+\frac{3}{2}+\frac{7}{2}\right)\left(x+\frac{3}{2}-\frac{7}{2}\right)\)
\(\Leftrightarrow3\left(x-2\right)\left(x+5\right)\)
d) \(x^2-9x+18\)
\(\Leftrightarrow x^2-3x-6x+18\)
\(\Leftrightarrow x\left(x-3\right)-6\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)
TK MK NKA !!!! TH@NK !!!
a) x2 - 3x + 2 ( như này mới phân tích được ạ :) )
= x2 - x - 2x + 2
= x( x - 1 ) - 2( x - 1 )
= ( x - 2 )( x - 1 )
b) x2 - 6x + 8
= x2 - 2x - 4x + 8
= x( x - 2 ) - 4( x - 2 )
= ( x - 4 )( x - 2 )
c) 3x2 + 9x - 30
= 3( x2 + 3x - 10 )
= 3( x2 - 2x + 5x - 10 )
= 3[ x( x - 2 ) + 5( x - 2 )]
= 3( x + 5 )( x - 2 )
d) x2 - 9x + 18
= x2 - 3x - 6x + 18
= x( x - 3 ) - 6( x - 3 )
= ( x - 6 )( x - 3 )
a, \(x^2-3x+2=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)=\left(x-2\right)\left(x-1\right)\)
b, \(x^2-6x+8=x^2-2x-4x+8\)
\(=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
c, \(3x^2+9x-30=3\left(x^2+3x-10\right)\)
\(=3\left(x^2-2x+5x-10\right)=3\left(x+5\right)\left(x-2\right)\)
