`a)1/[x-5x^2]-[25x-15]/[25x^2-1]`

`=[-(5x+1)-x(25x-15)]/[x(5x-1)(5x+1)]`

`=[-5x-1-25x^2+15x]/[x(5x-1)(5x+1)]`

`=[-25x^2+10x-1]/[x(5x-1)(5x+1)]`

`=[-(5x-1)^2]/[x(5x-1)(5x+1)]`

`=[1-5x]/[x(5x+1)]`

________________________________________________-

`b)(-1/[x^2-4x]+2/[16-x^2]-[-1]/[4x+16]):1/[4x]`

`=[-4(x+4)-8x+x(x-4)]/[4x(x-4)(x+4)].4x`

`=[-4x-16-8x+x^2-4x]/[(x-4)(x+4)]`

`=[x^2-16x-16]/[x^2-16]`

a, \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)

\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)

\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)

\(=\dfrac{5x-35}{5x\left(x-7\right)}\)

\(=\dfrac{5\left(x-7\right)}{5x\left(x-7\right)}=\dfrac{1}{x}\)

b, \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(5x-1)(5x+1)=25x²-7x+15

<=> 25x²-1-25x²+7x-15=0

<=>7x=16

=>x=16/7

(5x-1)(5x+1)=25x^2-7x+15

(5x-1)(5x+1)-25x^2+7x-15=0

(5x)^2-1^2-25x^2+7x-15=0

25x^2-1-25x^2+7x-15=0

7x-16=0

7x=16

x=16:7

x=16/7.vay x=16/7

a)\(dk,x\ne7;x\ne0\)

\(\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(7-x\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{5x\left(x-7\right)}=\frac{\left(4x+13\right)+\left(x-48\right)}{5x\left(x-7\right)}\\ \)

\(=\frac{5x-35}{5x\left(x-7\right)}=\frac{5\left(x-7\right)}{5x\left(x-7\right)}=\frac{1}{x}\)

b)

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

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

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

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

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

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

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

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

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

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

\(=\dfrac{1}{x-5x^2}+\dfrac{25x-15}{1-25x^2}\)

\(=\dfrac{1}{x\left(1-5x\right)}+\dfrac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\) MTC: \(x\left(1-5x\right)\left(1+5x\right)\)

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

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

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

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

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

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

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

b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

 Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)

  Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)

a)(x-1)(5x+3)=(3x-8)(x-1)

\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0

\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)

\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)