\(a,x^3=x\)

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

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

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

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

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

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 \(x=-\frac{3}{2}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

\(\Leftrightarrow a^2b+abc+a^2c+b^2a+b^2c+abc+bc^2+ac^2=0\)

\(\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

\(\Leftrightarrow....\)

Câu hỏi của Nguyễn Đa Vít - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo phần sau tại link trên.

\(a)A=(\frac{x}{(x+6)(x+6)}-\frac{x-6}{x(x+6)})\cdot\frac{x(x+6)}{2x-6}+\frac{x}{x-6}\)

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

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

\(b)\text{A luôn = -1 với mọi x}\)

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

                           \(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(b,-x^2+2x-4=-\left(x^2-2x+1+3\right)\)

                                    \(=-\left[\left(x-1\right)^2+3\right]< 0\forall x\)

d) \(x^4-5x^2+4\)

Đặt \(x^2=t\).Ta có:

\(x^4-5x^2+4=t^2-5t+4\)

\(t^2-t-4t+4=\left(t^2-t\right)-\left(4t-4\right)\)

\(=t\left(t-1\right)-4\left(t-1\right)=\left(t-1\right)\left(t-4\right)\)

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

a) \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x^2-4\right)\left(x-3\right)\)

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

b) \(2x^2-2y^2-6x-6y\)

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

\(=2\left(x+y\right)\left(x-y\right)-6\left(x+y\right)\)

\(=\left(x+y\right)\left[2\left(x-y\right)-6\right]\)

\(=\left(x+y\right)\left(2x-2y-6\right)\)

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

c) \(x^3+3x^2-3x-1\)

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

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

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

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

d) \(x^4-5x^2+4\)

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

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

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

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