\(\left(x-3\right)\left(x+3\right)=x^2-6x\\ \Leftrightarrow x^2-9=x^2-6x\\ \Leftrightarrow x^2-x^2+6x=9\\ \Leftrightarrow6x=9\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(\Leftrightarrow x^2-9=x^2-6x\\ \Leftrightarrow x^2-x^2-9+6x=0\\ \Leftrightarrow3.\left(2x-3\right)=0\\ \Leftrightarrow2x-3=0\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy S = \(\dfrac{3}{2}\)

\(\left(x-3\right)\left(x+3\right)=x^2-6x\\ \Leftrightarrow x^2-9=x^2-6x\\ \Leftrightarrow-6x=-9\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{3}{2}\right\}\)

Phương trình nào?

\(PT\Leftrightarrow\left(x^3+6x^2+11x-2\right)^2+13\left(x^3+6x^2+11x-2\right)+40=0\\ \Leftrightarrow\left(x^3+6x^2+11x-2+5\right)\left(x^3+6x^2+11x-2+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^3+6x^2+11x+3=0\left(\text{vô nghiệm}\right)\\x^3+6x^2+11x+6=0\end{matrix}\right.\\ \Leftrightarrow x^3+6x^2+11x+6=0\\ \Leftrightarrow x^3+x^2+5x^2+5x+6x+6=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+3x+6\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-3\end{matrix}\right.\)

\(x^4-6x^3+7x^2+6x-8=0\)

\(\Leftrightarrow x^4-4x^3-2x^3+8x^2-x^2+4x+2x-8=0\)

\(\Leftrightarrow x^3\left(x-4\right)-2x^2\left(x-4\right)-x\left(x-4\right)+2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^3-2x^2-x+2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left[x^2\left(x-2\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{-1;1;2;4\right\}\)

Vậy S={-1;1;2;4}

<=> x⁴+3x³+x²+3x³+9x²+3x+x²+3x+1=0

<=>x²(x²+3x+1)+3x(x²+3x+1)+(x²+3x+1)=0

<=> (x²+3x+1)(x²+3x+1)=0

<=>(x²+3x+1)²=0 => x²+3x+1=0 Giải PT bậc 2 để tìm x, bạn tự làm nốt nhé

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

\(x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0\)

\(\Leftrightarrow x^6-x^5-5x^5+5x^4+10x^4-10x^3-10x^3+10x^2+5x^2-5x-x+1=0\)

\(\Leftrightarrow x^5\left(x-1\right)-5x^4\left(x-1\right)+10x^3\left(x-1\right)-10x^2\left(x-1\right)+5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^5-5x^4+10x^3-10x^2+5x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^5-x^4-4x^4+4x^3+6x^3-6x^2-4x^2+4x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x-1\right)-4x^3\left(x-1\right)+6x^2\left(x-1\right)-4x\left(x-1\right)+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-4x^3+6x^2-4x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-x^3-3x^3+3x^2+3x^2-3x-x+1\right]=0\)

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

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

\(\Leftrightarrow\left(x-1\right)^4\left[x^2-2x+1\right]=0\Leftrightarrow\left(x-1\right)^6=0\Leftrightarrow x=1\)

b) http://olm.vn/hoi-dap/question/118763.html