bài tập toán

Question

Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng:

a) 5y7=20xy28x5y7=20xy28x;                              b) 3x(x+5)2(x+5)=3x23x(x+5)2(x+5)=3x2

c) x+2x−1=(x+2)(x+1)x2−1x+2x−1=(x+2)(x+1)x2−1;             d) x2−x−2x+1=x2−3x+2x−1x2−x−2x+1=x2−3x+2x−1

e) x3+8x2−2x+4=x+2x3+8x2−2x+4=x+2;

Answer 1

a) 5y.28x=140xy7.20xy=140xy} ⇒5y.28x=7.20xy

nên 5y7=20xy28x

b) 3x(x+5).2=3x.2(x+5)=6x(x+5)

nên 3x(x+5)2(x+5)=3x2

c) x+2x−1=(x+2)(x+1)x2−1

Vì (x+2)(x2−1)=(x+2)(x+1)(x−1).

d) 

(x2−x−2)(x−1)=x2.x+x2.(−1)+(−x).x+(−x).(−1)+(−2).x+(−2).(−1)=x3−x2−x2+x−2x+2=x3−2x2−x+2

(x+1)(x2−3x+2)=x.x2+x.(−3x)+x.2+1.x2+1.(−3x)+1.2=x3−3x2+2x+x2−3x+2=x3−2x2−x+2

⇒(x2−x−2)(x−1)=(x+1)(x2−3x+2)

Vậy  x2−x−2x+1=x2−3x+2x−1

e) x3+8x2−2x+4=x+2

Vì x3+8=x3+23=(x+2)(x2–2x+4)