Question

[Tạ Thu Hương]

Bài 5 : Phân tích các đa thức sau thành nhân tử : ( thêm bớt cùng một hạng tử )
a, x^4 + 4
b, x^4 + 64
c, x^8 + x^7 + 1
d, x^8 + x^4 + 1
e, x^5 + x + 1
f, x^3 + x^2 + 4
g, x^4 + 2x^2 - 24
h, x^3 - 2x - 4
i, a^4 + 4b^4
Giúp mk vs ạ mk cần gấp

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 5:

a) Ta có: \(x^4+4\)

\(=x^4+4\cdot x^2+4-4x^2\)

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

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

b) Ta có: \(x^4+64\)

\(=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

c) Ta có: \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+1\)

\(=x^6\left(x^2+x+1\right)-\left(x^6-1\right)\)

\(=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)

\(=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x-x^3-1\right)\)

d) Ta có: \(x^8+x^4+1\)

\(=x^8+x^4+x^6-x^6+1\)

\(=x^4\left(x^4+x^2+1\right)-\left(x^6-1\right)\)

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

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

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

g) Ta có: \(x^4+2x^2-24\)

\(=x^4+6x^2-4x^2-24\)

\(=x^2\left(x^2+6\right)-4\left(x^2+6\right)\)

\(=\left(x^2+6\right)\left(x^2-4\right)\)

\(=\left(x^2+6\right)\left(x-2\right)\left(x+2\right)\)

i) Ta có: \(a^4+4b^4\)

\(=a^4+4a^2b^2+4b^4-4a^2b^2\)

\(=\left(a^2+2b^2\right)^2-\left(2ab\right)^2\)

\(=\left(a^2-2ab+2b^2\right)\left(a^2+2ab+2b^2\right)\)

Answer 2

[Xuân Liệu]

ý e đâu