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\(x^2+7x+12\)
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:
\(x^2+7x+12\)
\(=\left(x^2+4x\right)+\left(3x+12\right)\)
\(=x\left(x+4\right)+3\left(x+4\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
Ta có : x4 + 8x2 + 7x + 8
= x4 - x + 8x2 + 8x + 8
= x(x3 - 1) + 8(x2 + x + 1)
= x(x - 1)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 - x)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 8)
Học tốt nhé !
a)\(x^3+4x^2-7x-10=x^3+x^2+3x^2+3x-10x-10=x^2\left(x+1\right)+3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x-10\right)=\left(x+1\right)\left[\left(x^2+5x\right)-\left(2x+10\right)\right]=\left(x+1\right)\left(x+5\right)\left(x-2\right)\)
b) \(x^8+x+1=x^8-x^2+x^2+x+1=x^2\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(2x^4+3x^3-7x^2-6x+8\)
\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)
\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+x^2-4x+4x^2+2x-8\right)\)
\(=\left(x-1\right)\left[x\left(2x^2+x-4\right)+2\left(2x^2+x-4\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(2x^2+x-4\right)\)
\(x^2-7x+12\)
\(=\left(x^2-4x\right)-\left(3x-12\right)\)
\(=x\left(x-4\right)-3\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
\(-2x^4-7x^3-x^2+7x+3\)
\(=-2x^3\left(x+1\right)-5x^2\left(x+1\right)+4x\left(x+1\right)+3\left(x+1\right)\)
\(=-\left(x+1\right)\left(2x^3+5x^2-4x-3\right)\)
\(=-\left(x+1\right)\left[2x^2\left(x-1\right)+7x\left(x-1\right)+3\left(x-1\right)\right]\)
\(=-\left(x+1\right)\left(x-1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(2x+1\right)\)
\(x^2\) - 7\(x\) - 8
= (\(x^2\) + \(x\)) - 8\(x\) - 8
= \(x\).(\(x\) + 1) - 8.(\(x\) + 1)
= (\(x+1\)).(\(x-8\))
x²-7x-8
x²-8x+x-8
x(x-8)+(x-8)
(x-8)(x+1)