Dùng quy tắc đổi dấu rồi thực hiện các phép tính
a) \(\frac{4x+13}{5x(x-7)}-\frac{3x+6}{5x(x-7)}\)
b) \(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}\)
Câu a:
\(\frac{4x+13}{5x(x-7)}-\frac{3x+6}{5x(x-7)}=\frac{4x+13}{5x(x-7)}+\frac{x-48}{-5x(7-x)}\)
\(\frac{4x+13}{5x(x-7)}+\frac{x-48}{5x(x-7)}=\frac{4x+13+x-48}{5x(x-7)}= \frac{5x-35}{5x(x-7)}\)
\(=\frac{5(x-7)}{5x(x-7)}=\frac{1}{x}\)
Câu b:
\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x(1-5x)}+\frac{25x-15}{-(25x^2-1)}\)
\(=\frac{1}{x(1-5x)}+\frac{25x-15}{1-25x^2}=\frac{1}{x(1-5x)}+\frac{25x-15}{(1-5x)(1+5x)}\)
\(=\frac{1+5x+x(25x-15)}{x(1-5x)(1+5x)}= \frac{1+5x+25x^2-15}{x(1-5x)(1+5x)}\)
\(=\frac{1-10x+25x^2}{x(1-5x)(1+5x)}=\frac{(1-5x)^2}{x(1-5x)(1+5x)} =\frac{1-5x}{x(1+5x)}\)
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