Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow x^3-3x^2-3x^2+9x-10x+30\)
\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-10\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)\left(x+2\right)\)
\(=\left(x^3-2x^2\right)+\left(8x^2-16x\right)+\left(15x-30\right)\)
\(=x^2\left(x-2\right)+8x\left(x-2\right)+15\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+8x+15\right)\)
\(=\left(x-2\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x-2\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x-2\right)\left(x+3\right)\left(x+5\right)\)
x3-6x2-x+30
=x3-5x2-x2+5x-6x+30
=(x-5)(x2-x-6)
=(x-5)(x-3)(x+2)
\(x^3-6x^2-x+30=x^3-3x^2-3x^2+9x-10x+30.\)
\(=x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-3x-10\right)\)
\(=\left(x-3\right)\left(x^2+2x-5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x+2\right)-5\left(x+2\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
Vậy \(x^3-6x^2-x+30=\left(x-3\right)\left(x+2\right)\left(x-5\right)\)
\(x^2+2x-3\)
\(=x^2-x+3x-3\)
\(=x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x+3\right)\)
\(2x^2+6x-x-3\)
\(=2x\left(x+3\right)-\left(x+3\right)\)
\(=\left(x+3\right)\left(2x-1\right)\)
a)\(x^2+2x-3=x^2+3x-x-3\)
\(=x\left(x+3\right)-\left(x+3\right)\)
\(=\left(x+3\right)\left(x-1\right)\)
a) \(x^{12}-3x^6+1\)
\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)
\(=\left(x^6-1\right)^2-\left(x^3\right)^2\)
\(=\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
c, \(2x^2+x-3=x\left(2x-3\right)\)
d, \(6x^2-x-15=x\left(6x-15\right)\)
TK MIK NHA
\(2x^2+x-3=2x^2-2x+3x-3=2x\left(x-1\right)+3\left(x-1\right)=\left(x-1\right)\left(2x+3\right) \)
a)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2=x^4+x^2+4-2x^3-4x+4x^2+x^2-4x+4\)
\(=x^4-2x^3+6x^2-8x+8=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)=\left(x^2-2x+2\right)\left(x^2+4\right)\)
b)\(x^4+6x^3+7x^2-6x+1=\left(x^2\right)^2+\left(3x\right)^2+\left(-1\right)^2+2.x^2.3x\)+2.3x.(-1)+2.x2.(-1)
\(=\left(x^2+3x-1\right)^2\)
x3 - 6x2 - x + 30
= (x + 2).x2 - 6x2 - x + 30/x + 2
= x2 - 8x + 15
= (x + 2)(x - 3)(x - 5)
\(x^3-6x^2-x+30\)
\(=\left(x^3-8x^2+15x\right)+\left(2x^2-16x+30\right)\)
\(=x\left(x^2-8x+15\right)+2\left(x^2-9x+15\right)\)
\(=\left(x^2-8x+15\right)\left(x+2\right)\)
\(=\left(x^2-3x-5x+15\right)\left(x+2\right)\)
\(=\left[x\left(x-3\right)-5\left(x-3\right)\right]\left(x+2\right)\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)
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) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
k trc làm sau sẵn kb lun
k trc làm sau đấy