a)

ĐKXĐ: \(x\ne-4\)

Để A nguyên thì \(3x+21⋮x+4\)

\(\Leftrightarrow3x+12+9⋮x+4\)

mà \(3x+12⋮x+4\)

nên \(9⋮x+4\)

\(\Leftrightarrow x+4\inƯ\left(9\right)\)

\(\Leftrightarrow x+4\in\left\{1;-1;3;-3;9;-9\right\}\)

\(\Leftrightarrow x\in\left\{-3;-5;-1;-7;5;-13\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{-3;-5;-1;-7;5;-13\right\}\)

b) ĐKXĐ: \(x\ne\dfrac{1}{2}\)

Để B nguyên thì \(2x^3-7x^2+7x+5⋮2x-1\)

\(\Leftrightarrow2x^3-x^2-6x^2+3x+4x-2+7⋮2x-1\)

\(\Leftrightarrow x^2\left(2x-1\right)-3x\left(2x-1\right)+2\left(2x-1\right)+7⋮2x-1\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-3x+2\right)+7⋮2x-1\)

mà \(\left(2x-1\right)\left(x^2-3x+2\right)⋮2x-1\)

nên \(7⋮2x-1\)

\(\Leftrightarrow2x-1\inƯ\left(7\right)\)

\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)

\(\Leftrightarrow2x\in\left\{2;0;8;-6\right\}\)

hay \(x\in\left\{1;0;4;-3\right\}\)(nhận)

Vậy: \(x\in\left\{1;0;4;-3\right\}\)

\(A=\left(2x+1\right)\left(x^2+1\right)+\dfrac{4}{2x+1}\) (chia đa thức)

Để A nguyên \(\Rightarrow4⋮2x+1\Rightarrow\left(2x+1\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x=\left\{-\dfrac{5}{2};-\dfrac{3}{2};-1;0;\dfrac{1}{2};\dfrac{3}{2}\right\}\)

x thỏa mãn đk đề bài là \(x=\left\{-1;0\right\}\)

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

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 <=> \(9⋮x+3\)

1) Xét rằng x > 7 <=> A < 0

Lại xét x < 7 thì mẫu là một số nguyên dương. P/s A có tử và mẫu đều là số dương, mà tử lại bất biến

A(max) <=> mẫu 7 - x nhỏ nhất <=> 7 - x = 1 => x = 7 - 1 = 6 <=> A = 1

Từ những điều trên thì A sẽ có GTLN khi và chỉ khi x = 6

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)

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

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

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2}{x^2-4}\)