\(=\dfrac{y}{x\left(y-5x\right)}+\dfrac{25x-15y}{\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y\left(y+5x\right)+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y^2+5xy+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y^2+30xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(\dfrac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}=\dfrac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}=\dfrac{y-5x}{x\left(y+5x\right)}\)

\(\frac{1}{x-5x^2}\)\(-\)\(\frac{25x-15}{25x^2-1}\)\(=\)\(\frac{-1}{x\left(5x-1\right)}\)\(-\)\(\frac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\)

\(=\)\(\frac{-1\left(5x+1\right)}{x\left(5x-1\right)\left(5x+1\right)}\)\(-\)\(\frac{\left(25x^{ }-15\right)x}{\left(5x-1\right)\left(5x+1\right)x}\)

\(=\) \(\frac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}\)\(=\)\(\frac{-25x^2+10x-1}{x\left(5x-1\right)\left(5x+1\right)}\)

\(=\)\(\frac{-\left(5x-1\right)^2}{x\left(5x-1\right)\left(5x+1\right)}\)\(=\)\(\frac{-5x+1}{5x^2+x}\)

\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}\)

\(=\frac{1}{x\left(1-5x\right)}-\frac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\)

\(=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(5x+1\right)}\)

\(=\frac{5x+1}{x\left(1-5x\right)\left(5x+1\right)}+\frac{\left(25x-15\right)x}{x\left(1-5x\right)\left(5x+1\right)}\)

\(=\frac{5x+1+25x^2+15x}{x\left(1-5x\right)\left(5x-1\right)}\)

\(=\frac{25x^2-10x+1}{x\left(1-5x\right)\left(5x+1\right)}\)

\(=\frac{\left(5x-1\right)^2}{x\left(1-5x\right)\left(5x+1\right)}\)

\(=\frac{\left(1-5x\right)^2}{x\left(1-5x\right)\left(5x+1\right)}\)

\(=\frac{1-5x}{x\left(5x+1\right)}\)

a, \(\frac{x+2y}{8x^2y^5}-\frac{3x^2+2}{12x^4y^4}\)

=\(\frac{\left(x+2y\right)3x^2}{24x^4y^5}-\frac{\left(3x^2+2\right)2y}{24x^4y^5}\)

=\(\frac{3x^3+6x^2y}{24x^4y^5}-\frac{6x^2y+4y}{24x^4y^5}\)

=\(\frac{3x^3+6x^2y-6x^2y-4y}{24x^4y^5}\)

=\(\frac{3x^3-4y}{24x^4y^5}\)

b,\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)

=\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{\left(15y-25x\right)x}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y^2+5xy}{x\left(y-5x\right)\left(y+5x\right)}-\frac{15xy-25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)

=\(\frac{y-5x}{x\left(y+5x\right)}\)

c,\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)

=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^3-x^2\right)+\left(2x-2\right)}\)

=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{\left(x+5\right)x}{x\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{4x-4-x^2+x}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)

=\(\frac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)

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

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

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

\(\frac{y}{xy-5x^2}-\frac{15x-25x}{y^2-25x^2}\)

ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\y\ne\pm5x\end{cases}}\)

\(=\frac{y}{x\left(y-5x\right)}-\frac{-10x}{\left(y-5x\right)\left(y+5x\right)}\)

\(=\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{-10xx}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\frac{y^2+5xy+10x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(\frac{y}{xy-5x^2}-\frac{-10x}{y^2-25x^2}=\frac{y^3-25x^2y}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}-\frac{-10x^2y+50x^3}{\left(y^2-25x^2\right)\left(xy-5x^2\right)}\)

\(=\frac{y^3-25x^2y+10x^2y-50x^3}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}=\frac{y^3-15x^2y-50x^3}{\left(xy-5x^2\right)\left(y^2-25x^2\right)}=\frac{y^3-50x^3}{x\left(y-5x\right)^2\left(y+5x\right)}\)

kha sdaif dòng mik xin phép trình bày bằng lời ạ :

a) tìm MTC rồi quy đồng lên làm bình thường ại , tử cộng tử mấu giữ nguyên 

b) cx vậy ạ tách mẫu tìm MTC rồi ....

~ hok tốt ~

a) \(\dfrac{y}{xy-5x^2}-\dfrac{15y-25x}{y^2-25x^2}=\dfrac{y}{x\left(y-5x\right)}-\dfrac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\dfrac{x\left(15y-25x\right)}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)

\(=\dfrac{y-5x}{x\left(y+5x\right)}\)

b: \(=\dfrac{2}{x+2y}-\dfrac{1}{2y-x}+\dfrac{4y}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\dfrac{2x-4y+x+2y+4y}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{3x+2y}{\left(x-2y\right)\left(x+2y\right)}\)

\(\frac{5x^2+y^2}{xy}-\frac{3x-2y}{xy}\)

\(=\frac{5x^2+y^2-3x-2y}{xy}\)

Tham khảo nhé~

3) \(\dfrac{y}{xy-5y^2}-\dfrac{15y-25x}{y^2-25x^2}\) MTC: \(\left(xy-5y^2\right)\left(y^2-25x^2\right)\)

\(=\dfrac{y\left(y^2-25x^2\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}-\dfrac{\left(xy-5y^2\right)\left(15y-25x\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)

\(=\dfrac{y\left(y^2-25x^2\right)-\left(xy-5y^2\right)\left(15y-25x\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)

\(=\dfrac{\left(y^3-25x^2y\right)-\left(15xy^2-25x^2y-75y^3+125xy^2\right)}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)

\(=\dfrac{y^3-25x^2y-15xy^2+25x^2y+75y^3-125xy^2}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)

\(=\dfrac{76y^3-140xy^2}{\left(xy-5y^2\right)\left(y^2-25x^2\right)}\)

4) \(\dfrac{4-2x+x^2}{2+x}-2-x\)

\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2+x}{1}\)

\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{\left(2+x\right)\left(2+x\right)}{2+x}\)

\(=\dfrac{4-2x+x^2-\left(2+x\right)\left(2+x\right)}{2+x}\)

\(=\dfrac{4-2x+x^2-\left(2+x\right)^2}{2+x}\)

\(=\dfrac{4-2x+x^2-\left(4+4x+x^2\right)}{2+x}\)

\(=\dfrac{4-2x+x^2-4-4x-x^2}{2+x}\)

\(=\dfrac{-6x}{2+x}\)