a, đk: \(x\ge0,x\ne9,x\ne4\)

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

\(=\dfrac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

b,\(Q< -1=>\dfrac{-1}{\sqrt{x}-3}+1< 0< =>\dfrac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(< =>\dfrac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)

\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left[{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\end{matrix}\right.\)\(< =>9< x< 16\)

c, \(=>2Q=\dfrac{-2}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\in Z\)

\(< =>\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)\(=>x\in\left\{16;4\right\}\)(loại 4)

=>x=16

a) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\) 

Ta có \(x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>9\\x>2\end{matrix}\right.\) \(\Leftrightarrow x>9\)

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

\(=\dfrac{\left(x-4\right)-\left(x-2\sqrt{x}-3\right)-\left(3\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-1}{\left(\sqrt{x}-3\right)}=\dfrac{1}{3-\sqrt{x}}\)

b) \(Q< -1\Leftrightarrow\dfrac{1}{3-\sqrt{x}}< -1\) \(\Leftrightarrow\dfrac{1}{3-\sqrt{x}}+1< 0\) \(\Leftrightarrow\dfrac{4-\sqrt{x}}{3-\sqrt{x}}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-\sqrt{x}>0\\3-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4-\sqrt{x}< 0\\3-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow9< x< 16\)

Vậy để \(Q< -1\) thì \(S=\left\{x/9< x< 16\right\}\)

c) \(2Q\in Z\Leftrightarrow\dfrac{2}{3-\sqrt{x}}\in Z\)

\(\Rightarrow3-\sqrt{x}\inƯ\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}3-\sqrt{x}=2\\3-\sqrt{x}=-2\\3-\sqrt{x}=1\\3-\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=25\\x=4\\x=16\end{matrix}\right.\)

Kết hợp với ĐKXĐ,ta có để \(2Q\in Z\) thì \(x\in\left\{16;25\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

Ta có: \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)

\(=\dfrac{x-4-x+2\sqrt{x}+2-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{-1}{\sqrt{x}-3}\)

c) Để 2Q là số nguyên thì \(-2⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1\right\}\)

325253737747⁸⁹⁰⁷⁶⁵⁴³ chuyển đổi sang STN là?

1, để \(\dfrac{2x+1}{x+3}\) là 1 số nguyên 

= > 2x + 1 chia hết cho x + 3 ( x thuộc Z và x \(\ne3\) )

= > 2 ( x + 3 ) - 5 chia hết cho x + 3 

=> -5 chia hết cho x + 3 

hay x + 3 thuộc Ư(-5 ) \(\in\left\{\pm1;\pm5\right\}\)

Đến đây em tự tìm các giá trị của x

2, Tương tự câu 1, x - 1 chia hết cho x + 5 ( x thuộc Z và x khác - 5 )

= > - 6 chia hết cho x + 5 

= > \(x+5\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

....

3,  ( x - 1 ) ( y - 3 ) = 7 

x,y thuộc Z = > x - 1 ; y - 3 thuộc Ư(7)

và ( x - 1 )( y - 3 ) = 7

( 1 ) \(\left\{{}\begin{matrix}x-1=1\\y-3=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=10\end{matrix}\right.\)

(2) \(\left\{{}\begin{matrix}x-1=7\\y-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)

( 3) \(\left\{{}\begin{matrix}x-1=-1\\y-3=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-4\end{matrix}\right.\)

( 4 ) \(\left\{{}\begin{matrix}x-1=-7\\y-3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=2\end{matrix}\right.\)

Từ ( 1 ) , ( 2 ) , ( 3 ) , ( 4 ) các cặp giá trị ( x,y ) nguyên cần tìm là ....

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

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

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

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

\(a,\Leftrightarrow7⋮x-1\Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\\ b,\Leftrightarrow\dfrac{x-1+2}{x-1}\in Z\Leftrightarrow1+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-1;0;2;3\right\}\)

a) để 7/x-1 thuộc Z 

=> (x-1) thuộc ước 7(+-1;+-7)

x-1  -1     1      -7      7

x      0     2       -6     8

a) \(\dfrac{7}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

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

b) \(\dfrac{x+1}{x-1}\in Z\Rightarrow\dfrac{x-1}{x-1}+\dfrac{2}{x-1}=1+\dfrac{2}{x-1}\in Z\)

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

\(\Rightarrow x\in\left\{2;0;3;-1\right\}\)

⇔−840≤x40≤−540

Mà x

\(\dfrac{-1}{5}\le\dfrac{x}{40}\le\dfrac{-1}{8}\)

\(\Leftrightarrow\dfrac{-8}{40}\le\dfrac{x}{40}\le\dfrac{-5}{40}\)

\(\Leftrightarrow-8\le x\le-5\)

Mà x\(\in Z\)

\(\Rightarrow x\in\left\{-8;-7;-6;-5\right\}\)

Vậy ...

\(\dfrac{-1}{5}\le\dfrac{x}{40}< \dfrac{-1}{8}\)

<=>\(\dfrac{-8}{40}\le\dfrac{x}{40}< \dfrac{-5}{40}\)

\(\Leftrightarrow-8\le x< 5\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{-8;-7;-6\right\}\)

Vậy

1.

\(a,Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(x+7\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

\(b,Q\in Z\Leftrightarrow\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}\in Z\\ \Leftrightarrow-1-\dfrac{8}{\sqrt{x}-2}\in Z\)

Mà \(-1\in Z\Leftrightarrow\dfrac{8}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow8⋮\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8,-4,-2,-1,1,2,4,8\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;1;3;4;6;10\right\}\)

Mà \(x\in Z\) và \(\sqrt{x}\ge0\)

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

Vậy \(x\in\left\{0;1;4\right\}\) thì \(Q\in Z\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{1}{xy+x+xyz}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{1}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{xyz}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz}{y+1+yz}+\dfrac{1}{y+yz+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+1}{y+1+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+xyz}{y+xyz+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{y\left(z+xz\right)}{y\left(1+xz+z\right)}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{z+xz+1}{xz+z+1}\)

\(A=1\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)⇔\(A=\dfrac{z}{1+xz+z}+\dfrac{xz}{z+1+xz}+\dfrac{1}{xz+z+1}\)(vì xyz=1)

⇔\(A=\dfrac{z+xz+1}{xz+z+1}\)⇔\(A=1\)

Xong rồi nè bn ơi 

\(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(=\dfrac{1}{\dfrac{1}{z}+\dfrac{1}{yz}+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1}{y}+z+1}\)

\(=\dfrac{1}{\dfrac{y+1+yz}{yz}}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1+zy+y}{y}}\)

\(=\dfrac{yz}{y+1+yz}+\dfrac{1}{yz+y+1}+\dfrac{y}{1+zy+y}=\dfrac{y+yz+1}{y+yz+1}=1\)

a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5

=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5

=(x-2)/(2x^2-5x+5)(x-1)