\(M=\dfrac{x^2}{x^2-3x}\left(x\ne0;x\ne3\right)\\ M=\dfrac{x^2}{x\left(x-3\right)}\\ M=\dfrac{x}{x-3}\)
\(N=\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}\left(x\ne\pm1\right)\\ N=\dfrac{x-1+3x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x}{\left(x+1\right)\left(x-1\right)}\)
Đúng 4
Bình luận (1)
a) M=\(\dfrac{x^2}{x^2-3x}\)=\(\dfrac{x.x}{x\left(x-3\right)}\)=\(\dfrac{x}{x-3}\)
b)\(\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}\)=\(\dfrac{2x^2+4x+1}{x^3+x^2}\)
Đúng 0
Bình luận (0)
\(M=\dfrac{x^2}{x^2-3x}=\dfrac{x}{x-3}\)
\(N=\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}=\dfrac{x^2-x+3x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+1}{x-1}\)
Đúng 0
Bình luận (0)