1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)

\(=0\)

a: ĐKXĐ: x<>-2

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

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

Để A là số nguyên thì 3⋮x+2

=>x+2∈{1;-1;3;-3}

=>x∈{-1;-3;1;-5}

b: \(B=\frac{\left(x+1\right)^2+\left(x-1\right)^2}{2x^2-1}\)

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

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

Để B là số nguyên thì \(3\vdots2x^2-1\)

=>\(2x^2-1\in\left\lbrace1;-1;3;-3\right\rbrace\)

=>\(2x^2\in\left\lbrace2;0;4;-2\right\rbrace\)

=>\(x^2\in\left\lbrace1;0\right\rbrace\)

=>x∈{0;1;-1}

c: C là số nguyên

=>2x-3⋮3x-2

=>6x-9⋮3x-2

=>6x-4-5⋮3x-2

=>-5⋮3x-2

=>3x-2∈{1;-1;5;-5}

=>3x∈{3;1;7;-3}

=>x∈{1;1/3;7/3;-1}

mà x là số nguyên

nên x∈{1;-1}

Lời giải:

$A=\frac{2011(2011+n)}{4022+n}$

Để $A$ nguyên thì: $2011(2011+n)\vdots 4022+n$

$\Rightarrow 2011^2+2011(n+4022)-2011.4022\vdots 4022+n$

$\Rightarrow 2011^2-2011.4022\vdots 4022+n$

$\Rightarrow 2011^2-2011^2.2\vdots 4022+n$

$\Rightarrow 2011^2\vdots 4022+n$

$\Rightarrow 4022+n\in\left\{\pm 1; \pm 2011; \pm 2011^2\right\}$

$\Rightarrow n\in \left\{-4023; -4021; -2011; -6033; 4040099; -4048143\right\}$

a, \(A=\dfrac{5n-4-4n+5}{n-3}=\dfrac{n+1}{n-3}=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

a.\(A=\dfrac{2n+1}{n-3}+\dfrac{3n-5}{n-3}-\dfrac{4n-5}{n-3}\)

\(A=\dfrac{2n+1+3n-5-4n+5}{n-3}\)

\(A=\dfrac{n+1}{n-3}\)

\(A=\dfrac{n-3}{n-3}+\dfrac{4}{n-3}\)

\(A=1+\dfrac{4}{n-3}\)

Để A nguyên thì \(\dfrac{4}{n-3}\in Z\) hay \(n-3\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

n-3=1 --> n=4

n-3=-1 --> n=2

n-3=2 --> n=5

n-3=-2 --> n=1

n-3=4 --> n=7

n-3=-4 --> n=-1

Vậy \(n=\left\{4;2;5;7;1;-1\right\}\) thì A nhận giá trị nguyên

b.hemm bt lèm:vv

Bài 2:

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

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

=x^2+x-x^2+x

=2x

a) \(P=\dfrac{x-1+4\left(\sqrt{x}+1\right)+1}{x-1}.\dfrac{x-1}{x+2\sqrt{x}}\)

\(=\dfrac{x+4\sqrt{x}+4}{x+2\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)

b) \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}=1+\dfrac{2}{\sqrt{x}}\in Z\)

Do \(\sqrt{x}>0\)

\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{1;2\right\}\)

\(\Rightarrow x\in\left\{1;4\right\}\)

\(A=1:\dfrac{2011+n-2011}{2011+n}=\dfrac{n+2011}{n}\)

Để A là số nguyên thì \(n\inƯ\left(2011\right)\)

hay \(n\in\left\{-1;1;2011;-2011\right\}\)