a/

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

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow x=1\)

b/

\(\Leftrightarrow2x^2-4x-2x^2-6x=0\)

\(\Leftrightarrow-10x=0\)

\(\Leftrightarrow x=0\)

c/

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

\(\Leftrightarrow3x\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{3}{2}\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(9y^2+30y+25\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(3y+5\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\3x+5=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=-\frac{5}{3}\)

d/

\(\Leftrightarrow4x^2-4x+1+4x^2+4x+1-2\left(4x^2-2x-2\right)+x=12\)

\(\Leftrightarrow8x^2+x+2-8x^2+4x+4=12\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)

Câu e) là: 2x³ + 6x² = x² + 3x nhé

a) \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)

\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\left(2x+5\right)^2=\left(x+2\right)^2\)

\(\Rightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)

\(\Rightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)

\(\Rightarrow\left(x+3\right)\left(3x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\3x=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

d) \(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

e) \(2x^3+6x^2=x^2+3x\)

\(\Rightarrow2x^3+6x^2-x^2-3x=0\)

\(\Rightarrow2x^3+5x^2-3x=0\)

\(\Rightarrow x\left(2x^2+5x-3\right)=0\)

\(\Rightarrow2x^2+5x-3=0\)

\(\Rightarrow2x^2-6x+x-3=0\)

\(\Rightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left(x^2-1\right)\left(x+2\right)-\left(x^3-8\right)-2x^2=0\)

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

\(\Rightarrow-x+10=0\)

\(\Rightarrow x=10\)

a) \(x^3+2x^2+2x+1=0\)

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

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

\(TH1:x+1=0\Leftrightarrow x=-1\)

\(TH2:x^2+x+1=0\)

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

Mà \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)nên loại TH2

Vậy x = 1

Câu a), x = -1 nha, kết luận nhầm

b) \(x^3-4x^2+12x-27=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+9\right)-4x\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-7x+9\right)=0\)

\(TH1:x-3=0\Leftrightarrow x=3\)

\(TH2:x^2-7x+9=0\)

\(\cdot\Delta=\left(-7\right)^2-4.9=13\)

Vậy pt của TH2 có 2 nghiệm phân biệt

\(x_1=\frac{7+\sqrt{13}}{2}\);\(x_2=\frac{7-\sqrt{13}}{2}\)

a) \(A=x^4-6x^3+11x^2-6x+1\)

\(A=\left(x^4-3x^3+x^2\right)-\left(3x^3-9x^2+3x\right)+x^2-3x+1\)

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

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

b) \(B=x^4-x^3+2x^2-11x-5\)

\(B=x^2\left(x^2-2x-1\right)+x\left(x^2-2x-1\right)+5\left(x^2-2x-1\right)\)

\(B=\left(x^2-2x-1\right)\left(x^2+x+5\right)\)

a)\(x^2+3x+6=x^2+2.\frac{3}{2}x+\frac{9}{4}+\frac{15}{4}=0\)

  \(\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\)

      \(\left(x+\frac{3}{2}\right)^2=-\frac{15}{4}\)

             Vì bình phương luôn lớn hơn hoặc bằng 0

                    Nên PT vô nghiệm

b)\(x^2-2x-3=0\)

   \(x^2-3x+x-3=0\)

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

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

d)\(x^3-2x^2-x+2=0\)

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

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

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

              x - 2 = 0                   x=2

c)\(2x^2+7x+3=0\)

    \(2x^2+x+6x+3=0\)

    \(x\left(2x+1\right)+3\left(2x+1\right)=0\)

     \(\left(2x+1\right)\left(x+3\right)=0\)

          \(\Rightarrow\orbr{\begin{cases}2x+1=0\\x+3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=-3\end{cases}}\)