Answer:

\(x^2-25y^2+2x+1\)

\(=(x^2+2x+1)-(5y)^2\)

\(=(x+1)^2-(5y)^2\)

\(=(x+1-5y)(x+1+5y)\)

Answer:

\(\frac{1}{x+1}+\frac{1}{x^2+3x-4}\)

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

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

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

\(\frac{1}{m+2}+\frac{1}{m^2+3m+2}+\frac{1}{4m^2+7m+3}\)

\(=\frac{\left(m+1\right)\left(4m+3\right)}{\left(m+2\right)\left(m+1\right)\left(4m+3\right)}+\frac{4m+3}{\left(m+2\right)\left(m+1\right)\left(4m+3\right)}+\frac{m+2}{\left(m+2\right)\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{\left(m+1\right)\left(4m+3\right)+4m+3+m+2}{\left(m+2\right)\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{4m^2+12m+8}{\left(m+2\right)\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{4}{4m+3}\)

\(\frac{1}{1-x}+\frac{2x}{x^2+1}\)

\(\text{MTC:}\left(x-1\right)\left(x+1\right)\)

\(\frac{1}{1-x}=\frac{-1}{x-1}=\frac{-1.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{-x-1}{\left(x-1\right)\left(x+1\right)}\)

\(\frac{2x}{x^2-1}=\frac{2x}{\left(x-1\right)\left(x+1\right)}\)

\(\frac{-1}{x-1}+\frac{2x}{x^2-1}=\frac{\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}=\frac{-x+1+2x}{\left(x-1\right)\left(x+1\right)}\)

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

                                                                                                                    \(=\frac{1}{x-1}\)

\(\text{Hok tốt!}\)

\(\text{@Kaito Kid}\)

Có gì sai cho em xin lỗi,tại mới học lớp 7;(((

bài nay mình ko biết làm đúng ko nữa

đương nhiên là em ko lm đc rùi