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\(A=x^3-x^2-8x+12\)
\(=x^3-2x^2+x^2-2x-6x+12\)
hay \(A=x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x+6\right)\)
\(=\left(x+2\right)^2\left(x+3\right)\)
\(A=x^3-x^2-8x+12\)
\(=x^3-2x^2+x^2-2x-6x+12\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-6\right)\)
\(=\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(=\left(x-2\right)^2\left(x+3\right)\)
Chúc bạn học tốt.
\(2x^3+x^2-4x-12=2x^3-4x^2+5x^2-10x+6x-12\)
\(=2x^2\left(x-2\right)+5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+5x+3\right)\)
\(=\left(x-2\right)\left[2x\left(x+1\right)+3\left(x+1\right)\right]\)
\(=\left(x-2\right)\left(x+1\right)\left(2x+3\right)\)
Xin lỗi bạn, mình làm sai.
\(2x^3+x^2-4x-12=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)
x4 + 2x3 + 5x2 + 4x -12=0
<=> x4 - x3 + 3x3 - 3x2 + 8x2 - 8x + 12x - 12 = 0
<=> ( x4 - x3 ) + ( 3x3 - 3x2 ) + ( 8x2 - 8x ) + ( 12x - 12 ) = 0
<=> ( x - 1 ) ( x3 + 3x2+ 8x +12) = 0
<=> ( x -1 ).[ ( x3 + 2x2 ) + ( x2 + 2x ) + ( 6x +1) ] = 0
<=>( x - 1). ( x + 2 ).( x2 + x + 6 ) = 0
<=> x = 1 hoặc x = -2
\(x^2+7x+12\)
cách 1: \(=x^2+4x+3x+12\)
\(=x\left(x+4\right)+3\left(x+4\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
cách 2: \(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
cách 3: \(=\left(x^2+7x+12,25\right)-0.25\)
\(=\left(x+3.5\right)^2-0.5^2\)
\(=\left(x+3.5+0.5\right)\left(x+3.5-0.5\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
lấy đâu ra 8 cách vậy trời!!!!!!!!!!!!!!!
Cách 1:
\(x^2+7x+12\)
\(=\left(x^2+4x\right)+\left(3x+12\right)\)
\(=x\left(x+4\right)+3\left(x+4\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
cậu ơi cái này vô nghiệm mà bậc 2 thì lằm sao mà phân tích được
\(x^2-x-12\)
\(=x^2+3x-4x-12\)
\(=x\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
mình đổi dấu nha!
\(x^6-64x^{12}=\left(x^3\right)^2-\left(8x^6\right)^2=\left(x^3-8x^6\right)\left(x^3+8x^6\right).\)
\(=x^6\left(1-8x^3\right)\left(1+8x^3\right)=x^6\left(1-2x\right)\left(1+2x+4x^2\right)\left(1+2x\right)\left(1-2x+4x^2\right)\)
\(\times^2+7\times+12\)
\(=(\times^2+4\times)+\left(3\times+12\right)\)
\(=\times\left(\times+4\right)+3\left(\times+4\right)\)
\(=\left(\times+4\right)\left(\times+3\right)\)
\(x^2+7x+12=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
Trả lời:
a) \(x^3+2x=x\left(x^2+2\right)\)
b) \(3x^3-12x^2=3x^2\left(x-4\right)\)
\(x^2-9+x-3=\left(x-3\right)\left(x+3\right)+\left(x-3\right)=\left(x-3\right)\left(x+3+1\right)=\left(x-3\right)\left(x+4\right)\)
\(x^3-x-12\)
\(=x^3-4x+3x-12\)
\(=x\left(x^2-4\right)+3\left(x-4\right)\)
\(=x\left(x-4\right)\left(x+4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left[x\left(x+4\right)+3\right]\)
\(=\left(x-4\right)\left(x^2+4x+3\right)\)
hic bạn ơi chỗ x^2 - 4 phải tách thành (x - 2)(x + 2) ạ