Question

câu 1: phân tích đa thức sau thành nhân tử:

a/ \(x^2-y^2+5x-5y\)

b/ \(x^2+4x+4\)

c/\(\left(x-3\right).\left(x+3\right)-\left(x-3\right)^2\)

câu 2: làm tính chia:

\(\left(x^4-2x^3+4x^2-8x\right).\left(x^2+4\right)\)

câu 3: chứng minh rằng: \(x^2-2x+2\)>0 với mọi x

Answer 1

Câu 1:

a) \(x^2-y^2+5x-5y=\left(x^2-y^2\right)+\left(5x-5y\right)=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)=\left(x-y\right)\left(x+y+5\right)\)

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

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

Bài 2: Đặt phép chia ra

Bài 3: Có: \(x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\ge1>0\left(đpcm\right)\)

Answer 2

1

a, \(x^2-y^2+5x-5y=\left(x^2-y^2\right)+\left(5x-5y\right)=\left(x+y\right)\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)+5\left(x+y\right)\)b.

\(x^2+4x+4=x^2+2x2+2^2=\left(x+2\right)^2\)

c.\(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2=x^2-3^2-\left(x^2-6x+3^2\right)=x^2-9-x^2+6x+9=6x\)

Chả biết đúng hay sai =))

Answer 3

Câu 1 : a,\(x^2-y^2+5x-5y=\left(x^2-y^2\right)+\left(5x-5y\right)\)

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

b,\(x^2+4x+4=\left(x+2\right)^2\)

c,\(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2\)

=\(\left(x-3\right)\left(x+3-x+3\right)\)

=\(\left(x-3\right).6\)