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\(x^3-3x^2+3x-1\) =0
=>\(\left(x-1\right)^3\)=0
=>x-1=0
=>x=1
vậy x =1
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
x2-2x-3 = x2-3x+x-3 = x(x-3) + (x-3) = (x+1)(x-3)
CHọn mình nha :)
\(x^2-2x-3\)
\(=x^2+x-3x-3\)
\(=x\left(x+1\right)-3\left(x+1\right)\)
\(=\left(x+1\right)\left(x-3\right)\)
hk tot
^^
a, \(x^3+x^2-x+2=x^3+2x^2-x^2-2x+x+2\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+x+2\)
\(=\left(x+2\right)\left(x^2-x+1\right)\)
b, \(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=\left(x+2\right)\left(x^2-8x+15\right)\)
a: Sửa đề: x^3-x^2+5x-5
=x^2(x-1)+5(x-1)
=(x-1)(x^2+5)
b: x^3+4x^2+x-6
=x^3-x^2+5x^2-5x+6x-6
=(x-1)(x^2+5x+6)
=(x-1)(x+2)(x+3)
c: \(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
i don't now
mong thông cảm !
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\(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\) \(-2\)
\(=\) \(x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x^2+x+1\right)\left(x-2\right)\)
\(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\) \(\left(x^2-8x+15\right)\left(x+2\right)\)
\(=\left(x^2-5x-3x+15\right)\left(x+2\right)\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)