a: ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)

b: \(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\left(\dfrac{10x-25}{x\left(x+5\right)}-\dfrac{x}{x-5}\right)\)

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

\(=\dfrac{10x-25}{10x^2-50x-25x+125-x^3-5x^2}\)

\(=\dfrac{10x-25}{-x^3+5x^2-75x+125}\)

a) ĐKXĐ : \(x\ne\pm5,x\ne0,x\ne\frac{5}{2}\)

Rút gọn :

Ta có : \(P=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{5\left(2x-5\right)}{x\left(x+5\right)}+\frac{x}{5-x}\)

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

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

\(=\frac{1}{x-5}-\frac{x}{x-5}=\frac{1-x}{x-5}\)

Vậy : \(P=\frac{1-x}{x-5}\) với \(x\ne\pm5,x\ne0,x\ne\frac{5}{2}\)

b) Để \(P=2013\Leftrightarrow\frac{1-x}{x-5}=2013\)

\(\Leftrightarrow\frac{1-x}{x-5}-2013=0\)

\(\Leftrightarrow\frac{1-x-2013\left(x-5\right)}{x-5}=0\)

\(\Rightarrow10066-2014x=0\)

\(\Leftrightarrow2014x=10066\)

\(\Leftrightarrow x=\frac{10066}{2014}\approx4,999\)( thỏa mãn )

c) Để P là số nguyên \(\Leftrightarrow1-x⋮x-5\)

\(\Leftrightarrow-\left(x-5\right)-4⋮x-5\)

\(\Leftrightarrow4⋮x-5\)

\(\Leftrightarrow x-5\inƯ\left(4\right)\)

\(\Leftrightarrow x-5\in\left\{-1,1,-2,2,-4,4\right\}\)

\(\Leftrightarrow x\in\left\{4,6,3,7,1,9\right\}\) ( thỏa mãn ĐKXĐ và \(x\inℤ\) )

Vậy \(x\in\left\{4,6,3,7,1,9\right\}\) để P là số nguyên .

ĐKXĐ : \(x^2-5x\ne0\Leftrightarrow x\left(x-5\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne5\end{cases}}\)

a) \(A=\frac{x^2-10x+25}{x^2-5x}\)

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

\(A=\frac{x-5}{x}\)

b) Để phân thức bằng 0 thì \(x-5=0\Leftrightarrow x=5\)

Mà ĐKXĐ \(x\ne5\)=> ko có giá trị của x để phân thức bằng 0

c) Để phân thức bằng 0 thì :

\(\frac{x-5}{x}=\frac{5}{2}\)

\(2x-10=5x\)

\(-10=3x\)

\(x=\frac{-3}{10}\)

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

b,Để phân thức có giá trị bằng 0 thì \(\frac{x-5}{x}=0\)

Mà: Theo điều kiện ta có: \(x\ne0\)

nên để: \(\frac{x-5}{x}=0\)thì: \(x-5=0\Leftrightarrow x=5\)

c,Để phân thức có giá trị bằng 5/2 thì:

\(\frac{x-5}{x}=\frac{5}{2}\)

\(\Leftrightarrow2\left(x-5\right)=5x\)

\(\Leftrightarrow2x-10=5x\)

\(\Leftrightarrow2x-5x=10\)

\(\Leftrightarrow-3x=10\Rightarrow x=-\frac{10}{3}\)

=.= hk tốt!!

\(a,\)\(đkxđ\)của \(A\)\(:\)\(\hept{\begin{cases}x^2-25\ne0\\x^2+5x\ne0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)\left(x+5\right)\ne0\\x\left(x+5\right)\ne0\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x\ne\pm5\\x\ne0\end{cases}}\)

\(đkxđ\)của \(B\)\(:\)\(\hept{\begin{cases}x^2+5x\ne0\\5-x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\left(x+5\right)\ne0\\5-x\ne0\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x\ne\pm5\\x\ne0\end{cases}}\)

\(b,\)\(A=\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}=\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\)

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

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

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

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

\(=\frac{-x^3+20x-25}{x\left(x-5\right)\left(x+5\right)}\)

\(\Rightarrow P=A:B=\frac{10x-25}{x\left(x+5\right)\left(x-5\right)}:\frac{x^3+20x-25}{x\left(x+5\right)\left(x-5\right)}\)

\(=\frac{10x-25}{x^3+20x-25}\)

Đề có vấn đề ko vậy babe -.- \(x^3+20x-25\)vẫn phân tích được, nhưng ko rút gọn được -.-

Lí do mk ko lm đc là ở chỗ đó đó

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

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

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

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...