Question

Bài 6. Tính:

a) ( x mũ 3 - 1 ) : ( x mũ 2 + x + 1 ).

b) ( x mũ 2 - 4x + 4 ) : ( x - 2 ).

Answer 1

\(a.\frac{x^3-1}{x^2+x+1}=\frac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)

\(b.\frac{x^2-4x+4}{x-2}=\frac{\left(x-2\right)^2}{x-2}=x-2\)

Answer 2

a) ( 𝑥 3 − 1 ) : ( 𝑥 2 + 𝑥 + 1 ) (x 3 −1):(x 2 +x+1) Ta nhận thấy: 𝑥 3 − 1 = ( 𝑥 − 1 ) ( 𝑥 2 + 𝑥 + 1 ) x 3 −1=(x−1)(x 2 +x+1) Vì vậy: 𝑥 3 − 1 𝑥 2 + 𝑥 + 1 = ( 𝑥 − 1 ) ( 𝑥 2 + 𝑥 + 1 ) 𝑥 2 + 𝑥 + 1 = 𝑥 − 1 x 2 +x+1 x 3 −1 ​ = x 2 +x+1 (x−1)(x 2 +x+1) ​ =x−1 Kết quả: 𝑥 − 1 x−1 ​ b) ( 𝑥 2 − 4 𝑥 + 4 ) : ( 𝑥 − 2 ) (x 2 −4x+4):(x−2) Ta nhận thấy: 𝑥 2 − 4 𝑥 + 4 = ( 𝑥 − 2 ) 2 x 2 −4x+4=(x−2) 2 Vì vậy: 𝑥 2 − 4 𝑥 + 4 𝑥 − 2 = ( 𝑥 − 2 ) 2 𝑥 − 2 = 𝑥 − 2 x−2 x 2 −4x+4 ​ = x−2 (x−2) 2 ​ =x−2 Kết quả: 𝑥 − 2 x−2 ​