Question

Cho phân thức: \(A=\frac{x^4-2x^2+1}{x^3-3x-2}\)

1. Tìm điều kiện của  \(x\) để A có nghĩa.

2. Rút gọn A.

3. Tính x để A < 1.

Answer 1

1. Để A có nghĩa thì \(x^3-3x-2\ne0\)

\(\Rightarrow\left(x^3-x\right)-\left(2x-2\right)\ne0\)

\(\Rightarrow x\left(x^2-1\right)-2\left(x-1\right)\ne0\)

\(\Rightarrow x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\ne0\)

\(\Rightarrow\left(x^2+x-2\right)\left(x-1\right)\ne0\)

\(\Rightarrow\left(x^2-1+x-1\right)\left(x-1\right)\ne0\)

\(\Rightarrow\left[\left(x+1\right)\left(x-1\right)+\left(x-1\right)\right]\left(x-1\right)\ne0\)

\(\Rightarrow\left(x-1\right)^2\left(x+2\right)\ne0\)

\(\Rightarrow x\ne1;x\ne-2\)

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

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

3/ Để A < 1 \(\Leftrightarrow\frac{\left(x+1\right)^2}{x+2}< 1\Leftrightarrow\left(x+1\right)^2< x+2\)

                                                        \(\Leftrightarrow x^2+2x+1< x+2\)

                                                         \(\Leftrightarrow x^2+x< 1\)

                                                           \(\Leftrightarrow x.\left(x+1\right)< 1\)

Vậy .....

Answer 2

1. A có nghĩa khi \(x^3-3x-2\ne0\)

\(\Leftrightarrow x^3+x^2-x^2-x-2x-2\ne0\)

\(\Leftrightarrow x^2\left(x+1\right)-x\left(x+1\right)-2\left(x+1\right)\ne0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x-2\right)\ne0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x-2x-2\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow x-2\ne0\)(do \(\left(x+1\right)^2\ge0\)) \(\Leftrightarrow x\ne2\)

2. Ta có :

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

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

=\(\left(x-1\right)\left(x^3+x^2-x-1\right)=\left(x-1\right)\left[x^2\left(x+1\right)-x\left(x+1\right)\right]\)

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

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

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

3. \(A< 1\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}< 1\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Leftrightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)

\(\Leftrightarrow\frac{x^2-3x+3}{x-2}< 0\)ta có \(x^2-3x+3=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{3}{4}=\left(x-\frac{3}{4}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\)(1) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\)(Thỏa mãn)

Vậy x<2 thì A<1