Question

phân tích đa thức thành nhân tử

1) 4x⁸ + 1

2) x⁷ +x⁵ +1

3) x⁷ +x⁵ -1

4) (x²-3x+5)² - 7x(x²-3x+5)+12x²

5) ab(a+b)+bc(b+c)+ac(a+c)+3abc

Answer 1

1) Ta có: \(4x^8+1=\left(4x^8+4x^4+1\right)-4x^4\)

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

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

Answer 2

4) Ta có: \(\left(x^2-3x+5\right)^2-7x\left(x^2-3x+5\right)+12x^2\) (1)

Đặt \(x^2-3x+5=a\) => (1) trở thành:

\(a^2-7ax+12x^2\) = \(\left(a^2-3ax\right)-\left(4ax-12x^2\right)\)

= \(a\left(a-3x\right)-4x\left(a-3x\right)\)

= \(\left(a-3x\right)\left(a-4x\right)\) (2)

Thay a = \(x^2-3x+5\) vào 2 ta được:

\(\left(x^2-6x+5\right)\left(x^2-7x+5\right)\)

= \(\left[\left(x^2-x\right)-\left(5x-5\right)\right]\left(x^2-7x+5\right)\)

= \(\left[x\left(x-1\right)-5\left(x-1\right)\right]\left(x^2-7x+5\right)\)

= \(\left(x-5\right)\left(x-1\right)\left(x^2-7x+5\right)\)

Answer 3

Làm lại câu 1:

Ta có: \(4x^8+1\) = \(\left(4x^8+4x^4+1\right)-4x^4\)

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

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

Answer 4

5) Ta có: \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(a+c\right)+3abc\)

= \(\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]\) \(+\left[ac\left(a+c\right)+abc\right]\)

= \(ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)

= \(\left(a+b+c\right)\left(ab+bc+ac\right)\)

Answer 5

2);3) https://hoc24.vn/hoi-dap/question/277483.html?pos=927686

Answer 6

Có dạng \(x^{3k+1}+x^{3n+2}+1\) thì có nhân tử chung là\(x^2+x+1\)