Question

Rút gọn phân thức sau:

A = \(\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}\)

B = \(\dfrac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}\) với x < 0

Answer 1

A = \(\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+5\right)^2}{x^2+5x+5}=x^2+5x+5\)B = \(\dfrac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}\)với x < 0

Với x < 0 thì |x-1| = 1-x, |x| = -x, ta có:

\(\dfrac{1-x-x+x}{\left(x-1\right)\left(3x-1\right)}=\dfrac{1-x}{\left(x-1\right)\left(3x-1\right)}=\dfrac{x-1}{\left(x-1\right)\left(1-3x\right)}=\dfrac{1}{1-3x}\)

Answer 2

Mysterious PersonNguyễn Đình DũngNguyễn Thanh Hằng๖ۣۜĐặng♥๖ۣۜQuý

giúp!! 2h30 đi rồi