Question

Rút gọn biểu thức:

A= (21/x^2-9 - x-4/3-x - x-1/3+x): (1 - 1/x+3)

các bn giúp mik vs !

 

Answer 1

\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-x-12}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x+2}{x+3}\)

\(=\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{x+2}\)

\(=\frac{3x+6}{x-3}\times\frac{1}{x+2}=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\frac{3}{x-3}\)

Answer 2

\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(\frac{x+2}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+2}{x+3}\right)\)

\(=\left(\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+2}{x+3}\right)\)

\(=\frac{6+3x}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)