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\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
x4+7x3+14x2+14x+4
=x4+7x3+4x2+10x2+14x+4
=(x4+4x2+4)+(7x3+14x)+10x2
=(x2+2)2+7x(x2+2)+10x2
=(x2+2)2+2x(x2+2)+5x(x2+2)+10x2
=(x2+2)(x2+2+2x)+5x(x2+2+2x)
=(x2+2+2x)(x2+2+5x)
\(x^3-2x^2-5x+6\)
\(=\left(x^3-4x^2+3x\right)+\left(2x^2-8x+6\right)\)
\(=x\left(x^2-4x+3\right)+2\left(x^2-4x+3\right)\)
\(=\left(x+2\right)\left(x^2-4x+3\right)\)
1. \(x^3-2x-5x+6\)
\(\Leftrightarrow x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+x-2\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x-1\right)\)
2. \(x^3-7x^2+15x-9\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x^2-6x\right)+\left(9x-9\right)\)
\(\Leftrightarrow x^2\left(x-1\right)-6x\left(x-1\right)+9\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x\left(x-3\right)-3\left(x-3\right)\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)^2\)
Tìm x
x3-x2-14x+24=0
<=> x3-3x2+2x2-6x-8x+24=0
<=> x2(x-3)+2x(x-3)-8(x-3)=0
<=> (x-3)(x2+2x-8)=0
<=> (x-3)(x2-2x+4x-8)=0
<=>(x-3)(x-2)(x+4)=0
<=> x-3=0 hay x-2=0 hay x+4=0
<=> x=3 hay x=2 hay x=-4
S={3;2;-4}
x4+2.x3-13.x2-14x+24
=x3.(x+2)-13x2+12x-26x+24
=x3.(x+2)-x.(13x-12)-2.(13x-12)
=x3.(x+2)-(13x-12)(x+2)
=(x+2)(x3-13x+12)
=(x+2)(x3-x-12x+12)
=(x+2)[x.(x2-1)-12.(x-1)]
=(x+2)[x.(x-1)(x+1)-12.(x-1)]
=(x+2)(x-1)[x.(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x.(x-3)+4.(x-3)]
=(x+2)(x-1)(x-3)(x+4)
\(x^3+x^2+9x-10x^2-10x+25x+25\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+25\right)=\left(x+1\right)\left(x-5\right)^2\)
???
sao vậy bn?