Question

Phân tích đa thức thành nhân tử

a, \(15xy-20x^2\)

b, \(\left(x-4\right)^2-\left(y-5\right)^2\)

c, \(x^4-9x^3-x^2+9x\)

d, \(2x^2-7x+5\)

e, ( x+1) (x+3) (x+5) (x+7) +15

f, 4x⁸ +1

Answer 1

Biết làm mỗi 2 câu thôi -.-

a) \(15xy-20x^2=5x\left(3y-4x\right)\)

d) \(2x^2-7x+5=2x^2-2x-5x+5\)

\(=2x\left(x-1\right)-5\left(x-1\right)=\left(2x-5\right)\left(x-1\right)\)

Answer 2

a) \(15xy-20x^2\)

\(=5x\left(3y-4x\right)\)

b) \(\left(x-4\right)^2-\left(y-5\right)^2\)

\(=\left[x-4-\left(y-5\right)\right]\cdot\left(x-4+y-5\right)\)

\(=\left(x-4-y+5\right)\left(x-9+y\right)\)

\(=\left(x+1-y\right)\left(x-9+y\right)\)

c) \(x^4-9x^3-x^2+9x\)

\(=x\left(x^3-9x^2-x+9\right)\)

\(=x\cdot\left[x^2\cdot\left(x-9\right)-\left(x-9\right)\right]\)

\(=x\left(x^2-1\right)\left(x-9\right)\)

\(=x\left(x-1\right)\left(x+1\right)\left(x-9\right)\)

d) \(2x^2-7x+5\)

\(=2x^2-2x-5x+5\)

\(=2x\left(x-1\right)-5\left(x-1\right)\)

\(=\left(2x-5\right)\left(x-1\right)\)

Answer 3

a) \(15xy-20x^2=5.\left(3xy-4x^2\right)\)

b) \(\left(x-4\right)^2-\left(y-5\right)^2\)

= \(\left(x-4-y+5\right).\left(x-4+y-5\right)\)

= \(\left(x-y+1\right).\left(x+y-9\right)\)

c) \(x^4-9x^3-x^2+9x\)

= \(x^3.\left(x-9\right)-x.\left(x-9\right)\)

= \(\left(x-9\right).\left(x^3-x\right)=\left(x-9\right).x.\left(x^2-1\right)\)

= \(\left(x-9\right).x.\left(x-1\right).\left(x+1\right)\)

d) \(2x^2-7x+5\)

= \(2x^2-2x-5x+5\)

= \(2x.\left(x-1\right)-5.\left(x-1\right)\)

= \(\left(x-1\right).\left(2x-5\right)\)

e) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

= \(\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\) (*)

ĐẶT \(x^2+8x+7\) = a

Phương trình (*) \(\Leftrightarrow a\left(a+8\right)+15\)

= \(a^2+8a+15\)

= \(\left(a+3\right)\left(a+5\right)\)

\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)

= \(\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

= \(\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

Answer 4

f, W = \(4x^8+1\)

\(=\left(2x^4\right)^2+1\)

\(=\left(2x^4\right)^2+4.x^4+1-4x^4\)

\(=\left(2x^4+1\right)^2-\left(2x^2\right)^2\)

\(=\left(2x^4+1-2x^2\right)\left(2x^4+1+2x^2\right)\)

e,

N = \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(v=x^2+8x+11\left(1\right)\)

\(N=\left(v+4\right)\left(v-4\right)+15\)

\(=v^2-16+15=v^2-1=\left(v-1\right)\left(v+1\right)\)

Thay vào biểu thức

\(N=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+2x+6x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)