a)B =  \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)

= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)

= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)

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

b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)

Thay x = -4 vào B, ta có:

B = \(\dfrac{-4.3}{-4+3}=12\)

c) Để B = \(\dfrac{-3}{5}\)

<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)

<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)

d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên

<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)

a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)

\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)

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

b: 2x^2+7x+3=0

=>(2x+3)(x+2)=0

=>x=-3/2(loại) hoặc x=-2(nhận)

Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)

d: |B|<1

=>B>-1 và B<1

=>B+1>0 và B-1<0

=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)

a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)

\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1\right\}\)

b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)

\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)

\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)

\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)

\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)

\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)

hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

a: Để C là số nguyên thì \(3x^3+6x^2+3x+x^2+2x+1-2⋮x^2+2x+1\)

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

=>(x+1)^2=1 hoặc (x+1)^2=2

=>\(x\in\left\{0;-2;\sqrt{2}-1;-\sqrt{2}-1\right\}\)

b: Để D là số nguyên thì \(x^4+x^2+x^3+x-29⋮x^2+1\)

=>\(x^2+1\in\left\{1;-1;29;-29\right\}\)

=>x^2+1=1 hoặc x^2+1=29

=>\(x\in\left\{0;2\sqrt{7};-2\sqrt{7}\right\}\)

a) ta có: A=\(\frac{21x+3}{7x+1}=\frac{3\left(7x+1\right)}{7x+1}=3\)   với x khác -1/7

Vâỵ vs mọi gt trị của x thuộc Z (x khác -1/7) thì A mang gt nguyên

b)ta có: B=\(\frac{3x+2}{2x+3}\)  => 2B=\(\frac{3\left(2x+3\right)-5}{2x+3}=3-\frac{5}{2x+3}\)

để B có giá trị nguyên <=>2B có gt nguyên <=> \(\frac{5}{2x+3}\) có gt nguyên<=> 2x+3 là các ước nguyên của 5

Ư(5)={-5 ; -1 ; 1 ; 5}

ta có bảng:

Vậy với x={-4 ; -2 ; -1 ; 1} thì B nguyên

\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)

a: Để A là số nguyên thì

x^3-2x^2+4 chia hết cho x-2

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

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

b: Để B là số nguyên thì

\(3x^3-x^2-6x^2+2x+9x-3+2⋮3x-1\)

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

=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)