ĐKXĐ: \(x\ne1\)

\(A=\frac{5x+1}{x^3-1}-\frac{1-2x}{x^2+x+1}-\frac{2}{1-x}\)

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

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

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

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

\(A=\frac{4}{x-1}\left(x^2+x+1\ne0\right)\)

b, ĐỂ x nhân giá trị nguyên 

\(\Rightarrow4⋮x-1\)

\(\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Nếu : x - 1 = 1 => x = 2 

  x - 1 = -1 => x = 0 

x - 1 = 2 => x = 3 

.....

\(4x\left(x-2007\right)-\left(x-2007\right)=0\)

\(=\left(4x-1\right)\left(x-2007\right)=0\)

\(\orbr{\begin{cases}4x-1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\\x-2007=0\Rightarrow x=2007\end{cases}}\)

kl : x = 1/4 hoặc 2007

\(4x\left(x-2007\right)-x+2007=0\)

\(4x\left(x-2007\right)-\left(x-2007\right)=0\)

\(\left(x-2007\right)\left(4x-1\right)=0\)

\(\leftrightarrow\orbr{\begin{cases}x-2007=0\\4x-1=0\end{cases}}\leftrightarrow\orbr{\begin{cases}x=2007\\x=\frac{1}{4}\end{cases}}\)

vậy \(x\in\left\{2007;\frac{1}{4}\right\}\)

ta có : \(2x^2-x+1⋮2x+1\)

\(\Rightarrow\left(2x^2+x\right)-\left(2x-1\right)+2⋮2x+1\)

\(\Rightarrow x\left(2x+1\right)-\left(2x+1\right)+2⋮2x+1\)

\(\left(2x+1\right)\left(x-1\right)+2⋮2x+1\)

\(\Rightarrow2⋮2x+1\Rightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Nếu : 2x + 1 = 1 => x = 0  ( TM ) 

    2x + 1 = -1 => x = -1  ( TM ) 

  2x + 1 = 2 => x = 3/2 ( loại ) 

2x + 1 = -2 => x = -3/2  ( loại ) 

\(\Rightarrow x\in\left\{0;-1\right\}\)

\(6x^2-5x-3xy+10x\)

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

\(6x^2-5x-3xy+10x\)

\(=\left(6x^2-3xy\right)+\left(10x-5y\right)\)

\(=3x\left(2x-y\right)+5\left(2x-y\right)\)

\(=\left(3x+5\right)\left(2x-y\right)\)

\(x^3+2x^2+3x=0\)\(\Leftrightarrow x.\frac{x^3+2x^2+3x}{x}=0\)

\(\Leftrightarrow x\left(x^2+2x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+2x+3=0\end{cases}}\)

Ta sẽ c/m \(x^2+2x+3=0\) vô nghiệm.Thật vậy:

\(x^2+2x+3=\left(x+1\right)^2+2\ge2\forall x\)

Từ đó suy ra \(x^2+2x+3=0\) vô nghiệm.

Vậy : x = 0

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

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

\(\Rightarrow2x^2-3x+1=0\)

\(2x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-1\right)=0\)

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

Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

ý còn lại tham khảo bài tth

\(a^2-b^2+a+b=\left(a-b\right)\left(a+b\right)+a+b=\left(a-b+1\right)\left(a+b\right)\)

\(c,\left(a+9\right)^2-36a^2=\left(a+9\right)^2-\left(6a\right)^2\)

\(=\left(a+9-6a\right)\left(a+9+6a\right)\)

\(a,x^3=x\)

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

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

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

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

\(KL:x=0;x=1\)

c) \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(\frac{2x^3+3x^2+2x+3}{x+\frac{3}{2}}\right)=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(2x^2+2\right)=0\) (bạn tự thực hiện phép chia đa thức giúp mình)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{2}=0\\2x^2+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\2\left(x^2+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\left(C\right)\\x^2=-1\left(L\right)\end{cases}}\)

Vậy đa thức có nghiệm duy nhất \(x=-\frac{3}{2}\)