Question

Cho M = \(\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right)\): \(\frac{2-4x}{x+1}\) - \(\frac{3x-x^2+1}{3x}\)

a, Rút gọn M

b,Tính giá trị của M biết x = 2006

c, Tìm x để M < 0

d, Tìm x ϵ Z để \(\frac{1}{M}\)ϵ Z

Answer 1

Lời giải:
a)

ĐKXĐ: \(x\neq 0; x\neq - 1\)

\(M=\frac{(x+2)(x+1)+2.3x-3.3x(x+1)}{3x(x+1)}:\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-8x^2+2}{3x(x+1)}.\frac{x+1}{2-4x}-\frac{3x-x^2+1}{3x}=\frac{2(1-4x^2)}{3x(2-4x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{2(1-2x)(1+2x)}{6x(1-2x)}-\frac{3x-x^2+1}{3x}=\frac{1+2x}{3x}-\frac{3x-x^2+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

b)

Khi $x=2006\Rightarrow M=\frac{2006-1}{3}=\frac{2005}{3}$

c)

\(M< 0\Leftrightarrow \frac{x-1}{3}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ suy ra $x< 1; x\neq 0; x\neq -1$

d)

Để \(\frac{1}{M}=\frac{3}{x-1}\in\mathbb{Z}\) thì \(3\vdots x-1\)

\(\Rightarrow x-1\in\left\{\pm 1;\pm 3\right\}\)

\(\Rightarrow x\in\left\{0;2;-2;4\right\}\)

Kết hợp đkxđ suy ra $x\in\left\{-2;2;4\right\}$

Answer 2

Lời giải:
a)

ĐKXĐ: \(x\neq 0; x\neq - 1\)

\(M=\frac{(x+2)(x+1)+2.3x-3.3x(x+1)}{3x(x+1)}:\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-8x^2+2}{3x(x+1)}.\frac{x+1}{2-4x}-\frac{3x-x^2+1}{3x}=\frac{2(1-4x^2)}{3x(2-4x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{2(1-2x)(1+2x)}{6x(1-2x)}-\frac{3x-x^2+1}{3x}=\frac{1+2x}{3x}-\frac{3x-x^2+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

b)

Khi $x=2006\Rightarrow M=\frac{2006-1}{3}=\frac{2005}{3}$

c)

\(M< 0\Leftrightarrow \frac{x-1}{3}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ suy ra $x< 1; x\neq 0; x\neq -1$

d)

Để \(\frac{1}{M}=\frac{3}{x-1}\in\mathbb{Z}\) thì \(3\vdots x-1\)

\(\Rightarrow x-1\in\left\{\pm 1;\pm 3\right\}\)

\(\Rightarrow x\in\left\{0;2;-2;4\right\}\)

Kết hợp đkxđ suy ra $x\in\left\{-2;2;4\right\}$