\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)

\(\dfrac{25.9-25.17}{-8.80-8.10}=\dfrac{-200}{-720}=\dfrac{5}{18}\)

\(\dfrac{48.12-48.15}{-3.270-3.30}=\dfrac{-144}{-900}=\dfrac{4}{25}\)

a, Quy đồng

\(\dfrac{5}{18}=\dfrac{5\times25}{18\times25}=\dfrac{125}{450}\)

\(\dfrac{4}{25}=\dfrac{4.18}{25.18}=\dfrac{72}{450}\)

Xét \(x<4\Rightarrow |x-4|=4-x\)

                    \(|x-5|=5-x\)

Biểu thức \(A=4-x+5-x=9-2x\)

Xét \(4\leq x<5 \Rightarrow |x-4|=x-4\) và \(|x-5|=5-x\) thay vào \(A=1\)

Xét \(x\geq5\Rightarrow|x-4|=x-4\) và \(|x-5|=x-5\) thay vào \(A=2x-9\)

\(|x-5|\)luôn \(\ge0\)

\(\Rightarrow\hept{\begin{cases}|x-5|=x-5\\|x-5|=-\left(x-5\right)=-x+5\end{cases}}\)

\(|x-4|\)luôn \(\ge0\)

\(\Rightarrow\hept{\begin{cases}|x-4|=x-4\\|x-4|=-\left(x-4\right)=-x+4\end{cases}}\)

Ta có các trường hợp:

\(\hept{\begin{cases}\text{|x-5|+|x-4|}=\left(x-5\right)+\left(x-4\right)=x-5+x-4=2x-9\\\text{|x-5|+|x-4|}=\left(-x+5\right)+\left(x-4\right)=-x+5+x-4=1\end{cases}}\)

\(\hept{\begin{cases}\text{|x-5|+|x-4|}=\left(-x+4\right)+\left(x-5\right)=-x+4+x-5=-1\\\text{|x-5|+|x-4|}=\left(-x+4\right)+\left(-x+5\right)=-x+4-x-5=-2x-1\end{cases}}\)

a: Ta có: \(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{x+2}{x\sqrt{x}+1}\)

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

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

Tham Khảo:

a) 38.7

b) 108,875

Câu rút gọn là dụng ý nghệ thuật để tránh lặp, thừa ý.

Câu sai ngữ pháp là lỗi.