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: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

mình cảm ơn!

a: \(N=\dfrac{x+\sqrt{x}+1+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x+\sqrt{x}+2}{x\sqrt{x}-1}\)

b: \(P=M\cdot N\)

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

\(=\dfrac{3x+3\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)

Cái này mình chỉ rút gọn được P thôi, còn P nguyên thì mình xin lỗi bạn rất nhiều nha

uk

(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì 2n⋮n-2

2n-4+4⋮n-2

2n-4⋮n-2⇒4⋮n-2

n-2∈Ư(4)⇒Ư(4)={1;-1;2;-2;4;-4}

n∈{3;1;4;0;6;-2}

(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{48.50}\)

=\(\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+...+\dfrac{2}{48.50}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)

=\(\dfrac{3}{2}.\dfrac{2}{25}\)

=\(\dfrac{3}{25}\)

Giải:

(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì \(2n⋮n-2\) 

\(2n⋮n-2\) 

\(\Rightarrow2n-4+4⋮n-2\) 

\(\Rightarrow4⋮n-2\) 

\(\Rightarrow n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) 

Vậy \(n\in\left\{0;1;3;4;6\right\}\)

(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+\dfrac{3}{14.16}+...+\dfrac{3}{48.50}\) 

\(=\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{48.50}\right)\) 

\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\) 

\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\) 

\(=\dfrac{3}{2}.\dfrac{2}{25}\) 

\(=\dfrac{3}{25}\) 

Chúc bạn học tốt!

bài 1

để A∈Z

\(=>n+3\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(=>\left\{{}\begin{matrix}n+3=-1\\n+3=1\end{matrix}\right.=>\left\{{}\begin{matrix}n=-4\\n=-2\end{matrix}\right.\)

vậy \(n\in\left\{-4;-2\right\}\)  thì \(A\in Z\)

Để A nguyên

⇒ \(\left(n+3\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)

n+3        1           -2

n           -2           -4

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