\(3x\left(x+2\right)=\left(x+2\right)^2\\ \Leftrightarrow3x\left(x+2\right)-\left(x+2\right)^2=0\\ \Leftrightarrow\left(x+2\right)\left(3x-x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{-2;1\right\}\)

\(3x\left(x+2\right)=\left(x+2\right)^2\\ \Leftrightarrow3x\left(x+2\right)=\left(x+2\right)\left(x+2\right)\\ \Leftrightarrow3x\left(x+2\right)-\left(x+2\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(3x-\left(x+2\right)\right)=0\\ \Leftrightarrow\left(x+2\right)\left(3x-x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\2x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy : \(S=\left\{-2,1\right\}\)

I do not know what to do.

a) Ta có: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\left(1\right)\)

Với a = 4

Thay vào phương trình (t) ta được:

  \(\frac{x+2}{x+2}+\frac{x-2}{x-2}=2\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

\(\Leftrightarrow2x^2=2x^2-8\)

\(\Leftrightarrow0x=-8\)

Vậy phương trình vô nghiệm

b) Nếu x = -1

\(\Rightarrow\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)

\(\Leftrightarrow\frac{-1+a}{1}+\frac{-3}{-1-a}=2\)

\(\Leftrightarrow\frac{\left(-1+a\right)\left(-1-a\right)}{-1-a}+\frac{-3}{-1-a}=\frac{2\left(-1-a\right)}{-1-a}\)

\(\Leftrightarrow1+a-a-a^2-3=-2-2a\)

\(\Leftrightarrow-a^2+2a=-2-1+3\)

\(\Leftrightarrow a\left(2-a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=0\\2-a=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)

Vậy a = {0;2}

NĂM MỚI VUI VẺ

\(a,\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)

\(\frac{x+2+2}{x+2}+\frac{x-4+2}{x-4}=2\)

=> \(1+\frac{2}{x+2}+1+\frac{2}{x-4}=2\)

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

=> x=1 (t/m \(x\ne-2\) và \(x\ne4\))

Hình như đề của bạn sai nên mình sửa lại nhé

x⁴ + 2x³ +5x² +4x-12=0

⇔x⁴-x³+3x³-3x²+8x²-8x+12x-12=0

⇔x³(x-1)+3x²(x-1)+8x(x-1)+12(x-1)=0

⇔(x-1)(x³+3x²+8x+12)=0

⇔(x-1)(x+2)(x²+x+6)=0

ta có x²+x+6 >0 ∀x

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

Vậy...

Đề sai không bạn