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

Ta có: \(\left(2x-1\right)^3-\left(3x^2-1\right)\left(x-2\right)+\left(x+3\right)^3\)

\(=8x^3-12x^2+6x-1-\left(3x^3-6x^2-x+2\right)+x^3+9x^2+27x+27\)

\(=9x^3-3x^2+33x+26-3x^3+6x^2+x-2\)

\(=6x^3+3x^2+34x+24\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)

-x+1 nha

giup mik vs cac bn.

Đề bài sai rồi bạn ! Mình sửa :

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)

b) \(P=\left(\frac{x-1}{x+1}-\frac{x+1}{x-1}\right):\frac{2x}{3x-3}\)

\(\Leftrightarrow P=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)

\(\Leftrightarrow P=\frac{x^2-2x+1-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)

\(\Leftrightarrow P=\frac{-4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)

\(\Leftrightarrow P=\frac{-6}{x+1}\)

c) Để P nhận giá trị nguyên

\(\Leftrightarrow\frac{-6}{x+1}\inℤ\)

\(\Leftrightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)

Ta loại các giá trị ktm

\(\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)

Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)

a: ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

b: \(B=\dfrac{x+3-1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+2+1}{x+2}\)

\(=\dfrac{x+2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+2}=\dfrac{1}{x-3}\)

c: Để B nguyên thì \(x-3\in\left\{1;-1\right\}\)

hay \(x\in\left\{4;2\right\}\)

Lời giải:

$(x-1)^3-(x-1)(x^2+x+1)=(x-1)[(x-1)^2-(x^2+x+1)]=(x-1)(x^2-2x+1-x^2-x-1)=(x-1)(-3x)=-3x(x-1)$

căn bậc hai không có số âm

\(\sqrt{-1}\) đó

a) ĐK : x ≥ 0 ; x ≠ 1

A=\(\frac{x-\sqrt{x}}{\sqrt{x}-1}:\frac{x+\sqrt{x}}{\sqrt{x}+1}\)

=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

=\(\sqrt{x}:\sqrt{x}\)

=1

Vậy A=1 với x ≥ 0 ; x ≠ 1

b) Vì A=1 nên không thể thay x