Đố: Phân tích đa thức \(x^2-8x+12\) thành nhân tử bằng nhiều cách nhất!
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x2-8x + 16 - 4 = ( x - 4 )2-22= ( x -4-2 ) . ( x-4+2 ) = ( x - 6 ) .( x -2 )
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\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)
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\(12x^2+7x-12=12x^2-5x+12x-12\)
\(=x\left(12x-5\right)+12\left(x-1\right)\)
Đề sai rồi bạn ời
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\(\Leftrightarrow x^3-2x^2+x^2-2x-6x+12\)
\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x-6\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x-2x-6\right)\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2\left(x+3\right)\)
T I C K ủng hộ nha
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\(2x^3-5x^2+8x-3\)
\(\Leftrightarrow2x^3-x^2-4x^2+2x+6x+3\)
\(\Leftrightarrow x^2\cdot\left(2x-1\right)-2x\cdot\left(2x-1\right)+3\cdot\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-1\right)\cdot\left(x^2-2x+3\right)\)
\(2x^3-5x^2+8x-3\)
\(=2x^3-x^2-4x^2+2x+6x-3\)
\(=x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\)
C1
Ta có \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-6\right)\left(x-2\right)\)
C2
ta có \(x^2-8x+12\)
\(=x^2-8x+16-4\)
\(=\left(x-4\right)^2-2^2\)
\(=\left(x-4-2\right)\left(x-4+2\right)\)
\(=\left(x-6\right)\left(x-2\right)\)