ĐKXĐ: \(x\ne\pm3\)

a

Khi x = 1:

\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)

Khi x = 2:

\(A=\dfrac{3.2+2}{2-3}=-8\)

Khi x = \(\dfrac{5}{2}:\)

\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)

b

Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên

\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)

Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)

c

Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên

\(\Rightarrow x\left(x+3\right)-7⋮\left(x+3\right)\)

\(\Rightarrow-7⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)

d

\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)

=> Để A, B cùng là số nguyên thì x = 4.

Ta có \(A=\dfrac{4x-3}{x+2}=\dfrac{4x+8-11}{x+2}=4-\dfrac{11}{x+2}\)

Để \(A\) nguyên thì \(11⋮\left(x+2\right)\Rightarrow\left(x+2\right)\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+2=1\\x+2=-1\\x+2=11\\x+2=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=9\\x=-13\end{matrix}\right.\)

Vậy tất cả các x thỏa ycbt là x=-1;x=-3;x=9 hoặc x=-13

Để A là số nguyên thì \(4x-3⋮x+2\)

\(\Leftrightarrow-11⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-1;-3;9;-13\right\}\)

a: ĐKXĐ: x∉{0;3;-3}

\(\frac{9}{x^3-9x}+\frac{1}{x+3}\)

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

\(=\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)

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

\(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\)

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

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

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

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

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

b: A=2

=>\(-\frac{3}{x-3}=2\)

=>\(x-3=-\frac32\)

=>\(x=-\frac32+3=\frac32\) (nhận)

c: A nguyên

=>-3⋮x-3

=>x-3∈{1;-1;3;-3}

=>x∈{4;2;6;0}

Kết hợp ĐKXĐ, ta được: x∈{4;2;6}

Ý kiến mk thôi

đơn giản b = 6 ;a=6

Bài 4:

a. Ta thấy: $x^2-x+2=(x-\frac{1}{2})^2+1,75>0$ với mọi $x$.

Do đó để $B=\frac{x^2-x+2}{x-3}<0$ thì $x-3<0$

$\Leftrightarrow x<3$ 

b. 

$B=\frac{x(x-3)+2(x-3)+8}{x-3}=x+2+\frac{8}{x-3}$

Với $x$ nguyên, để $B$ nguyên thì $x-3$ phải là ước của 8.

$\Rightarrow x-3\in\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$

$\Rightarrow x\in \left\{4; 2; 5; 1; -1; 7; 11; -5\right\}$

Bài 5:

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

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