ta có : 2x^3=0 hoặc 6x^2=0 hay x-8=0

TH1:2x^3=0 => 2x^3 = 0^3 =>2x=0 => x=0/2 =>x=0

TH2:6X^2=0 =>6x^2=0^2 => 6x = 0 => x = 0/6 => x=0

TH3:x-8=0 => x = 0 + 8 => x = 8

      Vậy x  E ( 0 , 8)

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Xin chúc mừng bạn !

333 x 33 = 10989

=10989

chùi chăm chỉ quá

học tốt

kb

Bài 1 :

\(x^2\left(x-3\right)-4x+12=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{\pm2\right\}\end{cases}}}\)

Bài 2 :

\(x-1-x^2\)

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

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

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

Vì \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge0\forall x\)

\(\Rightarrow-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\le0\forall x\left(đpcm\right)\)

\(A=\frac{3}{2-x}+\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)

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

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

\(A=\frac{-3x-6+3x-6+3x^2}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{-12+3x^2}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(-4+x^2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(A=3\)

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

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

       \(=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right)\left(x+2\right)}\)

       \(=\frac{3x^2-6x}{\left(x-2\right)\left(x+2\right)}\)

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

      \(=\frac{3x}{x+2}\)

\(b,ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm2\\x\ne-1\end{cases}}}\)

Ta có : \(P=A:B=\frac{3x}{x+2}:\frac{x+1}{x+2}\)

                              \(=\frac{3x}{x+2}.\frac{x+2}{x+1}\)

                             \(=\frac{3x}{x+1}\)

                             \(=\frac{3x+3}{x+1}-\frac{3}{x+1}\)

                           \(=3-\frac{3}{x+1}\)

Để P nguyên thì \(3-\frac{3}{x+1}\inℤ\)

                          \(\Leftrightarrow\frac{3}{x+1}\inℤ\)

Vì \(x\inℤ\Rightarrow x+1\inℤ\)

Ta có bảng :

Vậy \(x\in\left\{-4;-2;0;2\right\}\)