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

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

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

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

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

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

\(P=\frac{5}{x-5}+\frac{x}{5-x}\)

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

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

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

\(P=-1\)

=> Giá trị của biểu thức P không phụ thuộc vào biến

                                                   đpcm

ĐKXĐ: x khác 4

Đặt A=\(\frac{x^3-16x}{8-2x}\)

A=\(\frac{x^3-16x}{8-2x}=\frac{x.\left(x^2-16\right)}{8-2x}\)

để A=0 => \(\orbr{\begin{cases}x=0\\x^2=16\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

\(4x^3=x\)

\(\Rightarrow4x^3-x=0\)

\(\Rightarrow x\left(4x^2-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\4x^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\Rightarrow x=\frac{1}{2}\end{cases}}\)

KL....

= 2^x - 8 + 4^x + 13 = 4^x + 2^x + 5
= 0