Ai giúp mình nhanh với nha

a: Để A là phân số thì n+5<>0

hay n<>-5

b: Để A=-1/2 thì n-1/n+5=-1/2

=>2n-2=-n-5

=>3n=-3

hay n=-1

c: Để A là số nguyên thì \(n-1⋮n+5\)

\(\Leftrightarrow n+5\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(n\in\left\{-4;-6;-3;-7;-2;-8;1;-11\right\}\)

Giúp mình những câu mik in đậm nhé

Thanks mọi người

a) ĐKXĐ: \(x\ge0,x\ne9\)

\(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)

b) \(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{64}+3}=\dfrac{3}{8+3}=\dfrac{3}{11}\)

c) \(2A=\dfrac{6}{\sqrt{x}+3}=1\Rightarrow\sqrt{x}+3=6\Rightarrow x=9\left(tm\right)\)

g) \(A=\dfrac{3}{\sqrt{x}+3}\in Z\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

h) \(A=\dfrac{3}{\sqrt{x}+3}\in Z\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

k) \(2A=\dfrac{6}{\sqrt{x}+3}=m\)

Help me

Ta có: \(A=\frac{6n-9+13}{2n-3}=\frac{3\left(2n-3\right)+13}{2n-3}\)

Mà: 3 ( 2n - 3 ) chia hết cho 2n - 3

=> 13 chia hết cho 2n - 3 => 2n - 3 E Ư(13) = {1,-1,13,-13}

=> 2n E {4,2,16,-10}

Ta có bảng sau:

ĐKXĐ: \(x\ge0;x\ne4\)

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

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

b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)

c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)

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

\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)

a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b: Thay x=36 vào A, ta được:

\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)

\(\Leftrightarrow4\sqrt{x}=2\)

hay \(x=\dfrac{1}{4}\)

d: Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

e: Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

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

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;4;0\right\}\)

hay \(x\in\left\{1;9;16;0\right\}\)

Lắm thế                                  

b: Để A là số nguyên thì 5n-9 chia hết cho 2n+4

=>10n-18 chia hét cho 2n+4

=>10n+20-38 chia hết cho 2n+4

=>\(2n+4\in\left\{1;-1;2;-2;19;-19;38;-38\right\}\)

=>\(n\in\left\{-\dfrac{3}{2};-\dfrac{5}{2};-1;-3;\dfrac{15}{2};-\dfrac{23}{2};17;-21\right\}\)

\(A=\frac{4x-3}{2x+1}=\frac{4x+2-5}{2x+1}=2-\frac{5}{2x+1}\)

a) A>0 => \(\frac{5}{2x+1}<2\Leftrightarrow2x+1>\frac{5}{2}\Leftrightarrow2x>\frac{3}{2}\Leftrightarrow x>\frac{3}{4}\)

b)A<0 => x <3/4 ; x khác -1/2

c)A =0 khi x = 3/4

d) A thuộc Z khi 2x+1 thuộc U(5) ={1;5;-1;-5}

2x+1 =1 => x =0

2x+1=-1 => x = -1

2x+1 =5 => x =2

2x+1 = -5 => x =-3

Để A là phân số thì 3n + 7 ko chia hết cho n + 1

<=> n + 1 khác Ư(4) = {-1;-2;-4;1;2;4}

=> n khác {-2;-3;-5;0;1;3}

Để A là số nguyên thì 3n + 7 chia hết cho n + 1

=> 3n + 3 + 4 chia hết cho n + 1

=> 3.(n + 1) + 4 chia hết cho n + 1

=>  4 chia hết cho n + 1

=> n + 1 thuộc Ư(4) = {-4;-2;-1;1;2;4}

=> n = {-5;-3;-2;0;1;3}

ko biết

d, ĐK:\(n+2\ne0\Leftrightarrow n\ne-2\)

\(e,A=2\\ \Leftrightarrow\dfrac{n+9}{n+2}=2\\ \Rightarrow n+9=2n+4\\ \Leftrightarrow n=5\\ A=4\\ \Leftrightarrow\dfrac{n+9}{n+2}=4\\ \Leftrightarrow n+9=4n+8\\ \Leftrightarrow3n=1\\ \Leftrightarrow n=\dfrac{1}{3}\)

\(f,A\in Z\\ \Rightarrow\dfrac{n+9}{n+2}\in Z\\ \Rightarrow\dfrac{n+2+7}{n+2}\in Z\\ \Rightarrow1+\dfrac{7}{n+2}\in Z\)

Để \(A\in Z\Rightarrow\dfrac{7}{n+2}\in Z\Rightarrow7⋮\left(n+2\right)\Rightarrow n+2\inƯ\left(7\right)\)

Ta có bảng:
 

Vậy \(n\in\left\{-9;-3;-1;5\right\}\)