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

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

\(A=\dfrac{x}{5x-1}\)

- \(x= 0,2\) là bị sai đó bạn. Nếu \(x= 0,2\) thì MT sẽ bằng \(0\).

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

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

\(A=\dfrac{x}{5x-1}\)

- Thay \(x=0,2\) vào ta được:

\(A=\dfrac{0,2}{5.0,2-1}=-\dfrac{1}{3}\)

\(\dfrac{1-\dfrac{7}{x}+\dfrac{10}{x^2}}{1-\dfrac{25}{x^2}}=\left(1-\dfrac{7}{x}+\dfrac{10}{x^2}\right):\left(1-\dfrac{25}{x^2}\right)\)

\(=\dfrac{x^3-7x^2+10x}{x^3}:\dfrac{x^2-25}{x^2}\)

\(=\dfrac{(x^3-7x^2+10x).x^2}{x^3.\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^5-7x^4+10x^3}{x^3.\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^3\left(x^2-7x+10\right)}{x^3\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2-7x+10}{\left(x-5\right)\left(x+5\right)}\)

\(Q=\dfrac{1}{x+1}-\dfrac{1}{x}\)

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

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

Thay \(x=1\) vào ta được:

\(Q=\dfrac{-1}{1\left(1+1\right)}=-\dfrac{1}{2}\)

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\)

\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{a}=k^4\)

\(\Rightarrow k=\pm1\)

- Với \(k=1\) :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}\)

\(\Rightarrow a=b=c=d\)

- Với \(k=-1\) :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=-1\)

\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-d\\d=-a\end{matrix}\right.\)

\(\Rightarrow a=-b=c=-d\)

\(\Rightarrow P=\dfrac{2a+a}{2a+a}+\dfrac{-2a-a}{-2a-a}+\dfrac{2a+a}{2a+a}+\dfrac{-2a-a}{-2a-a}\)

\(\Rightarrow P=4\)