a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x+2}{x+2}=1\)

b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)

\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)

\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)

\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)

\(1,\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

\(x-2\sqrt{x}-3\sqrt{x}+6\)

\(x-5\sqrt{x}+6\)

\(2,\left(x+2\right)\left(x-3\right)+x\left(x+1\right)\)

\(x^2+2x-3x-6+x^2+x\)

\(2x^2-6\)

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

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

a.\(\frac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) = \(\frac{5x+1-1+3x-2x^2+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\) =\(\frac{10x+2}{x^3-1}\)

b.\(\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{x^3+1}\)( đến đây dễ r đúng ko)

Dễ thế mak ko biết làm, đồ ngu như óc chó

Cứ làm như og giỏi lắm ế

Người ta thấy khó thì người ta hỏi

Có chết đâu

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

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

\(=x-2\sqrt{x}+\sqrt{x}-2\)

\(=x-\sqrt{x}-2\)

\(2.\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2\)

\(=x\left(x-2\right)+4\left(x-2\right)-\left(x^2-6x+9\right)\)

\(=x^2-2x+4x-8-x^2+6x-9\)

\(=8x-17\)

Trả lời:

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

2) \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2=x^2-2x+4x-8-\left(x^2-6x+9\right)\)\(=x^2+2x-8-x^2+6x-9=8x-17\)

3)  \(3x\left(2x^3-3x^2+5\right)=6x^4-9x^3+15x\)

Xin phép sửa đề:

Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}=\frac{x+3}{1-x^2}\) \(\left(x\ne\pm1\right)\)

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

\(\Rightarrow3x^2+4x+1-1+2x-x^2=-x^2-2x+3\)

\(\Leftrightarrow3x^2+8x-3=0\)

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

\(\Leftrightarrow3x\left(x+3\right)-\left(x+3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

Vậy tập nghiệm PT \(S=\left(-3;\frac{1}{3}\right)\)

Bài làm:

đk: \(x\ne-3;x\ne1\)

Ta có: \(\frac{x^2+6x+9}{1-x}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)

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

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

\(=-\frac{x^2-2x+1}{2x+6}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne1\end{cases}}\)

\(\frac{x^2+6x+9}{1-x}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x+3\right)^2}{x-1}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)