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

\(A=\left(\dfrac{x}{x+3}-\dfrac{2x}{3-x}+\dfrac{3x^2+9}{9-x^2}\right):\dfrac{3}{x-3}\)

\(=\left(\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{x-3}{3}\)

\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-9}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x-3}{3}\)

\(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x+3\right)}\cdot\dfrac{1}{3}\)

\(=\dfrac{3x-9}{3\left(x+3\right)}=\dfrac{x-3}{x+3}\)

b: Để P là số nguyên thì \(x-3⋮x+3\)

=>\(x+3-6⋮x+3\)

=>\(-6⋮x+3\)

=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{-2;-4;-1;-5;0;-6;-9\right\}\)

Lời giải:
a. ĐKXĐ: $x\neq \pm 3$

\(A=\left[\frac{x(x-3)}{(x+3)(x-3)}+\frac{2x(x+3)}{(x-3)(x+3)}-\frac{3x^2+9}{(x-3)(x+3)}\right].\frac{x-3}{3}\\ =\frac{x(x-3)+2x(x+3)-(3x^2+9)}{(x-3)(x+3)}.\frac{x-3}{3}\\ =\frac{3(x-3)}{(x-3)(x+3)}.\frac{x-3}{3}=\frac{x-3}{x+3}\)

b. 

$A=\frac{x-3}{x+3}=1-\frac{6}{x+3}$
Để $A$ nguyên thì $\frac{6}{x+3}$ nguyên.

Với $x$ nguyên, điều này xảy ra khi $6\vdots x+3$

$\Rightarrow x+3\in \left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$

$\Rightarrow x\in \left\{-2; -4; -1; -5; 0; -6; 3; -9\right\}$

Do $x\neq \pm 3$ nên $x\in \left\{-2; -4; -1; -5; 0; -6; -9\right\}$

a) ĐKXĐ: \(x\ne\pm3\)

\(A=\left(\dfrac{x}{x+3}-\dfrac{2x}{3-x}+\dfrac{3x^2+9}{9-x^2}\right):\dfrac{3}{x-3}\)

\(=\left(\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}\right)\cdot\dfrac{x-3}{3}\)

\(=\left[\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\right]\cdot\dfrac{x-3}{3}\)

\(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{3}\)

\(=\dfrac{3x-9}{x+3}\cdot\dfrac{1}{3}\)

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

Vậy \(A=\dfrac{x-3}{x+3}\) với \(x\ne\pm3\).

b) Ta có: \(A=\dfrac{x-3}{x+3}=\dfrac{x+3-6}{x+3}=1-\dfrac{6}{x+3}\)

Để \(A\) nhận giá trị nguyên thì \(\dfrac{6}{x+3}\) có giá trị nguyên

\(\Rightarrow6⋮x+3\)

\(\Rightarrow x+3\inƯ\left(6\right)\)

\(\Rightarrow x+3\in\left\{1;2;3;6;-1;-2;-3;-6\right\}\)

\(\Rightarrow x\in\left\{-2;-1;0;3;-4;-5;-6;-9\right\}\)

Kết hợp với ĐKXĐ của x, ta được: \(x\in\left\{-2;-1;0;-4;-5;-6;-9\right\}\)

\(\text{#}Toru\)

a) Để A xác định thì: \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\\\dfrac{x-1}{x+3}\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne1\end{matrix}\right.\)

Với \(x\ne\pm3;x\ne1\), ta có:

\(A=\left(\dfrac{2x}{x-3}+\dfrac{x}{x+3}+\dfrac{2x^2+3x+1}{9-x^2}\right):\dfrac{x-1}{x+3}\)

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

\(=\dfrac{2x^2+6x+x^2-3x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-1}\)

\(=\dfrac{x^2-1}{x-3}\cdot\dfrac{1}{x-1}\)

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

Vậy \(A=\dfrac{x+1}{x-3}\) với \(x\ne\pm3;x\ne1\).

b) Với \(x\ne\pm3;x\ne1\):

Để \(A=3\) thì \(\dfrac{x+1}{x-3}=3\)

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

\(\Leftrightarrow x+1=3x-9\)

\(\Leftrightarrow x-3x=-9-1\)

\(\Leftrightarrow-2x=-10\Leftrightarrow x=5\) (tm ĐKXĐ)

Vậy \(A=3\) tại \(x=5\).

c) Để \(A< 1\) thì \(\dfrac{x+1}{x-3}< 1\)

\(\Leftrightarrow\dfrac{x+1}{x-3}-1< 0\)

\(\Leftrightarrow\dfrac{x+1-\left(x-3\right)}{x-3}< 0\)

\(\Leftrightarrow\dfrac{4}{x-3}< 0\)

\(\Rightarrow x-3< 0\) (vì \(4>0\))

\(\Leftrightarrow x< 3\)

Kết hợp với ĐKXĐ của \(x\), ta được: \(x< 3;x\ne-3;x\ne1\)

Vậy \(A< 1\) khi \(x< 3;x\ne-3;x\ne1\).

\(\text{#}Toru\)

Lời giải:

a. ĐKXĐ: $x\neq -1; x\neq \pm 2$

\(A=\left[\frac{4x}{x(x+2)}+\frac{2}{x-2}+\frac{6-5x}{(x-2)(x+2)}\right].\frac{x-2}{x+1}\\ =\left[\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{(x-2)(x+2)}\right].\frac{x-2}{x+1}\\ =\frac{4(x-2)+2(x+2)+(6-5x)}{(x-2)(x+2)}.\frac{x-2}{x+1}\\ =\frac{x+2}{(x-2)(x+2)}.\frac{x-2}{x+1}=\frac{1}{x+1}\)

b.

$x^2-2x=8$

$\Leftrightarrow x^2-2x-8=0$

$\Leftrightarrow (x+2)(x-4)=0$

$\Leftrightarrow x=-2$ hoặc $x=4$. Do $x\neq -2$ nên $x=4$
Khi đó: $A=\frac{1}{x+1}=\frac{1}{4+1}=\frac{1}{5}$

c.

Với $x$ nguyên, để $A=\frac{1}{x+1}$ nguyên thì $1\vdots x+1$

$\Rightarrow x+1\in \left\{\pm 1\right\}$

$\Rightarrow x\in \left\{0; -2\right\}$

Do $x\neq -2$ nên $x=0$

a) MTC: \(3\left(x-4\right)\left(x+4\right)\)

+, \(\dfrac{-7}{x-4}=\dfrac{-21\left(x+4\right)}{3\left(x-4\right)\left(x+4\right)}\)

+, \(\dfrac{3}{3x+12}=\dfrac{3}{3\left(x+4\right)}=\dfrac{3\left(x-4\right)}{3\left(x-4\right)\left(x+4\right)}\)

+, \(\dfrac{-5}{16-x^2}=\dfrac{5}{x^2-16}=\dfrac{5}{\left(x-4\right)\left(x+4\right)}=\dfrac{15}{3\left(x-4\right)\left(x+4\right)}\)

Nhưng cái \(16-x^2\) thì đổi làm sao thành \(x-4\) vậy b

Biểu thức B không có giá trị nhỏ nhất bạn nhé. Bạn xem lại đề..