Question

Giúp m ình với, mình cảm ơn nhiều

Phân tích đa thức thành nhân tử

a, x² 16-4xy +4y²

b, x⁵-x⁴+x³-x²

c, x⁵+x³-x²-1

d, x⁴-3x³-x+3

Answer 1

a)x^2.16-4xy+4y^2

<=>16.x^2-2x2y+(2y)^2

<=>16(x-2y)^2

b)x^5-x^4+x^3-x^2

<=>(x^5-x^4)+(x^3-x^2)

<=>x^4(x-1)+x^2(x-1)

<=>(x-1)(x^4+x^2)

c)x^5+x^3-x^2-1

<=>(x^5+x^3)-(x^2+1)

<=>x^3(x^2+1)-(x^2+1)

<=>(x^2+1)(x^3-1)

d)x^4-3x^3-x+3

<=>(x^4-3x^3)-(x-3)

<=>x^3(x-3)-(x_3)

<=>(x-3)(x^3-1)

Answer 2

\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)

 

Answer 3

a)x².16-4xy+4y²

=(4x)²-3.x.2y+(2y)²

=(4x-2y)²

b)x5-x⁴+x³-x²

=(x5-x⁴)+(x³-x²)

=x⁴(x-1)+x²(x-1)

=(x⁴+x²).(x-1)

=x²(x²+1)(x-1)

c)x5+x³-x²-1

=(x5+x³)-(x²+1)

=x³(x²+1)-(x²+1)

=(x³-1)(x²+1)

d)x⁴-3x³-x+3

=(x⁴-3x³)-(x-3)

=x³(x-3)-(x-3)

=(x³-1)(x-3)