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a, \(x^3+4x^2-29x+24\)
\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left[x\left(x-3\right)+8\left(x-3\right)\right]\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
\(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
Chúc bạn học tốt.
\(x^3+4x^2-29x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]\)
\(=\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0
=> x3 + 4x2 - 29x + 24 = x2(x-1) + 5x(x-1) - 24(x-1)
= (x-1)(x2+5x-24) = (x-1)(x-3)(x+8)
b) ...
a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
a,\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left(x^2-3x+8x-24\right)\)
\(=\left(x-1\right)\left(x\left(x-3\right)+8\left(x-3\right)\right)\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
a x2 - 7x - 6 = x2 -6x-x-6
= x(x-6)+(x-6)
= (x+1)(x+6)
b x3 + 4x2 - 7x -10
=x3 - 2x2 + 6x2 -12x+5x-10
= x2(x-2) + 6x(x-2) + 5(x-2)
= (x2 +6x+5)(x-2)
=(x2 +x+5x+5)(x-2)
=[x(x+1)+5(x+1)](x-2)
=(x+5)(x+1)(x-2)
c x3 +x2-2
=x3-x2+2x2-2x+2x-2
=x2(x-1)+2x(x-1)+2(x-1)
=(x2+2x+2)(x-1)
d x2 +5x+6
=x2 +2x+3x+6
=x(x+2)+3(x+2)
=(x+3)(x-2)
Con 1 y e nhung luoi lam
mk chỉnh lại đề
\(x^2-7x+6\)
\(=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x-6\right)\)
\(d, x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-x-3x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]\)
\(=\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
\(e, x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+3\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
Mik ko giỏi p/t bậc 3 cho lắm nên có sai thì mong bạn sửa giúp :
d ) \(x^3+4x^2-29x+24\)
\(=x^2\left(x-1\right)+5x^2-10x+5-19x+19\)
\(=x^2\left(x-1\right)+5\left(x-1\right)^2-19\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-5-19\right)\)
\(=\left(x-1\right)\left(x^2+5x-14\right)\)
\(=\left(x-1\right)\left(x^2+7x-2x-14\right)\)
\(=\left(x-1\right)\left(x+7\right)\left(x-2\right)\)
e ) \(x^3+6x^2+11x+6\)
\(=x^2\left(x+1\right)+5x^2+10x+5+x+1\)
\(=x^2\left(x+1\right)+5\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left(x^2+5x+5+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)