a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)

\(Q=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\left(đk:x\ge0,x\ne4\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{8}{\sqrt{x}-2}=1+\dfrac{8}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Do \(x\ge0,x\ne4\)

\(\Rightarrow x\in\left\{0;1;9;16;36;100\right\}\)

Đkxđ: x # 4

Q = 1 + 8/(sqrt(x) - 2)

Q nguyên --> sqrt(x) - 2 là ước của 8

Do sqrt(x) >=0 nên sqrt(x) - 2 >= -2

TH1: sqrt(x) - 2 = -2 <=> x = 0 (thỏa)

TH2: sqrt(x) - 2 = -1 <=> x = 1 (thỏa)

Th3: sqrt(x) - 2 = 1 <=> x = 9(thỏa)

TH4: sqrt(x) - 2 = 2<=> x = 16 (thỏa)

Th5: sqrt(x) - 2 = 4 <=> x = 36 (thỏa)

Th6: sqrt(x) - 2 = 8 <=> x = 100 (thỏa)

a) \(Q=\) \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne1\right)\)

\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\) 

\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(Q=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(Q=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(Q=\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{2}{x-1}\)  \(\left(đpcm\right)\).

b) Để \(Q\in Z\) <=> \(\dfrac{2}{x-1}\in Z\) <=> \(x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

Vậy để biểu thức Q nhận giá trị nguyên thì \(x\in\left\{2;3\right\}\) 

1: Sửa đề: \(B=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

2: Để B<=-1/2 thì B+1/2<=0

=>-3/căn x+3+1/2<=0

=>-6+căn x+3<=0

=>căn x<=3

=>0<x<9

3: Để B là số nguyên thì \(\sqrt{x}+3=3\)

=>x=0

\(\Leftrightarrow\sqrt{x}-2\in\left\{-1;1;5\right\}\)

hay \(x\in\left\{1;9;49\right\}\)

Để P nguyên thì \(2\sqrt{x}-1⋮\sqrt{x}+1\)

\(\Leftrightarrow-3⋮\sqrt{x}+1\)

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

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

\(\Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\)

hay \(x\in\left\{0;4\right\}\)

ĐK: x>0,x\(\ne4\)

a) Ta thay x=\(\dfrac{1}{4}\) vào \(A=\dfrac{6}{x+2\sqrt{x}}=\dfrac{6}{\dfrac{1}{4}+2\sqrt{\dfrac{1}{4}}}=\dfrac{6}{\dfrac{1}{4}+2.\dfrac{1}{2}}=\dfrac{6}{\dfrac{1}{4}+1}=6:\left(\dfrac{1}{4}+1\right)=6:\dfrac{5}{4}=6.\dfrac{4}{5}=\dfrac{24}{5}=4,8\)B=\(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{4-x}\)

b) Ta có M=\(\dfrac{A}{B}=A\div B=\dfrac{6}{x+2\sqrt{x}}\div\dfrac{6}{4-x}=\dfrac{6}{x+2\sqrt{x}}.\dfrac{4-x}{6}=\dfrac{4-x}{x+2\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{2-\sqrt{x}}{\sqrt{x}}\)

Ta lại có M>1\(\Leftrightarrow\dfrac{2-\sqrt{x}}{\sqrt{x}}>1\Leftrightarrow2-\sqrt{x}>\sqrt{x}\Leftrightarrow2>2\sqrt{x}\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

Kết hợp với ĐK

Vậy 0<x<1 thì M>1

c) Ta có M\(=\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{2}{\sqrt{x}}-1\)

Vậy để \(M\in Z\) thì \(\sqrt{x}\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)

Vì \(\sqrt{x}>0\)

Nên \(\sqrt{x}\in\left\{1;2\right\}\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)

Vậy x=1 thì M\(\in Z\)

Nguyễn Việt LâmTrầNguyễn Thị Khánh Như Trương NgọcThảo Vyn Trung NguyênBonkingsaint suppapong udomkaewkanjanaPhạm TiếnKHUÊ VŨMysterious PersonThiên Hàn

a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)

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

hay \(x\in\left\{9;1;16;0\right\}\)

b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)

=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)

=>căn x+3=7

=>căn x=4

=>x=16

c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)

\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)

=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)

=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)

=>x=0 hoặc x=4

Mysterious Person giup mk nha

a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)

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

hay \(x\in\left\{9;1;16;0\right\}\)

b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)

=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)

=>căn x+3=7

=>căn x=4

=>x=16

c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)

\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)

=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)

=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)

=>x=0 hoặc x=4