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Phân tích đa thức thành nhân tử :
a) \(2,5x^2+3,7x+1,2\)
\(=2,5x^2+2,5x+1,2x+1,2\)
\(=\left(2,5x^2+2,5x\right)+\left(1,2x+1,2\right)\)
\(=2,5x\left(x+1\right)+1,2\left(x+1\right)\)
\(=\left(2,5x+1,2\right)\left(x+1\right)\)
b) \(4x^3+6x^2+9x+7\)
\(=4x^3+4x^2+2x^2+2x+7x+7\)
\(=\left(4x^3+4x^2\right)+\left(2x^2+2x\right)+\left(7x+7\right)\)
\(=4x^2\left(x+1\right)+2x\left(x+1\right)+7\left(x+1\right)\)
\(=\left(4x^2+2x+7\right)\left(x+1\right)\)
a) \(x^2-6x+8\)
\(=x^2-2\cdot x\cdot3+3^2-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
Còn lại tương tự
a) \(x^2-6x+8=x^2-2x-4x+8\)
\(=\left(x^2-2x\right)-\left(4x-8\right)\)
=x(x-2)-4(x-2) = (x-2)(x-4)
a) \(4x^2-4x+3=4x^2-4x+1+2\)
\(=\left(2x-1\right)^2+2>0\)\(\forall x\)
=> ko phân tích thành nhân tử được
b) \(9x^2+6x-8=9x^2+12x-6x-8\)
\(=3x\left(3x+4\right)-2\left(3x+4\right)=\left(3x-2\right)\left(3x+4\right)\)
c) \(3x^2-8x+4=3x^2-6x-2x+4\)
\(=3x\left(x-2\right)-2\left(x-2\right)=\left(3x-2\right)\left(x-2\right)\)
a) 12x3 + 4x2 + 9x + 3 = 4x2(3x + 1) + 3(3x + 1) = (4x2 + 3)(3x + 1)
b) x3 + 2x2 - x - 2 = x2(x + 2) - (x + 2) = (x2 - 1(x + 2) = (x - 1)(x + 1)(x + 2)
c) a3 + (a - b)3 = (a + a - b)[a2 - a(a - b) + (a - b)2] = (2a - b)(a2 - a2 + ab + a2 - 2ab + b2)
= (2a - b)(a2 - ab + b2)
a) 12x3 + 4x2 + 9x + 3
= 4x2(3x + 1) + 3(3x + 1)
= (4x2 + 3)(3x + 1)
b) x3 + 2x2 - x - 2
= x2(x + 2) - (x + 2)
= (x2 - 1)(x + 2)
c) a3 + (a - b)3
= a3 - a2(a - b) + a(a - b)2 + (a - b)a2 - (a - b)2a + (a - b)3
= a[(a2 - a(a - b) + (a - b)2] + (a - b)[a2 - a(a - b) + (a - b)2]
= (a + a - b)[(a2 - a(a - b) + (a - b)2]
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) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
x3 + 7x - 6=x2 . x + 7x - 22 + 2 = (x2 - 22) + (x+7x)+2=(x-2) . (x+2) + 8x + 2
x3 - 5x + 8x - 4=x2 . x -5x + 8x -22 = (x2 - 22) . (x -5x + 8x )=(x-2) . (x+2) . 4x
x3 - 9x2 + 6x + 16=x2 . x - 9x2 + 6x + 16 = (x2 - 9x2) . (x+6x) + 16=(x-9x) . (x+9x) . 7x + 16
k mk nha
b) \(9x^3+6x^2+x\)
\(=x\left(9x^2+6x+1\right)\)
\(=x\left(3x+1\right)^2\)
c) \(x^4+5x^3+15x-9\)
\(=\left(x^4-9\right)+5x\left(x^2+3\right)\)
\(=\left(x^2-3\right)\left(x^2+3\right)+5x\left(x^2+3\right)\)
\(=\left(x^2+3\right)\left(x^2-3+5x\right)\)
a) \(x^2-y^2+10y-25\)
\(=x^2-\left(y^2-10y+25\right)\)
\(=x^2-\left(y-5\right)^2\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
a) (x2-4x+3)(x2-10x+24)+8=((x2-x)-(3x-3))((x2-6x)-(4x-24))+8
=(x(x-1)-3(x-1))(x(x-6)-4(x-6))+8=(x-1)(x-3)(x-4)(x-6)+8=((x-1)(x-6))(x-3)(x-4))+8
=(x2-7x+6)(x2-7x+12)+8
Đặt x2-7x+6=a
Ta có : a(a+6)+8=a2+6a+8=(a+2)(a+4)=(x2-7x+8)(x2-7x+10)=(x2-7x+8)(x-5)(x-2)
b) Tương tự như câu a kết quả là (x-3)(x3+9x2+21x+9)
c) x4+x3+6x2+3x+9=(x4+x3+3x2)+(3x2+3x+9)=x2(x2+x+3)+3(x2+x+3)=(x2+x+3)(x2+2)
\(x^3+9x^2+6x-16\)
\(=x^3+x^2-2x+8x^2+8x-16\)
\(=x\left(x^2+x-2\right)+8\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x+8\right)\)
\(=\left(x^2-x+2x-2\right)\left(x+8\right)\)
\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]\left(x+8\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x+8\right)\)
AI nhanh k cho ạ :)) Cần hơi gấp ạ :))
Thank lắm ạ :))