Câu hỏi đâu bn??

đấy mà

1) Xét rằng x > 7 <=> A < 0

Lại xét x < 7 thì mẫu là một số nguyên dương. P/s A có tử và mẫu đều là số dương, mà tử lại bất biến

A(max) <=> mẫu 7 - x nhỏ nhất <=> 7 - x = 1 => x = 7 - 1 = 6 <=> A = 1

Từ những điều trên thì A sẽ có GTLN khi và chỉ khi x = 6

\(A=\frac{x+5}{x-4}=\frac{x-4+9}{x-4}=1+\frac{9}{x-4}\)

\(a)\)

\(\text{Để A có giá trị nguyên: }\)

\(\frac{9}{x-4}\in Z\)

\(x-4\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

\(\rightarrow x\in\left\{1;3;\pm5;7;13\right\}\)

\(b)\)

\(\text{Để A có giá trị lớn nhất: }\)

\(\frac{9}{x-4}\)\(\text{lớn nhất}\)

\(x-4=1\)

\(x=5\)

\(c)\)

\(\text{Để A đạt giá trị nhỏ nhất:}\)

\(\frac{9}{x-4}\)\(\text{nhỏ nhất}\)

\(x-4=-1\)

\(x=3\)

Cho \(A=\frac{x+5}{x-4}=\frac{x-4+9}{x-4}=\frac{x-4}{x-4}+\frac{9}{x-4}=1+\frac{9}{x-4}\left(ĐK:x\in Z,x\ne4\right)\)

Để A nguyên \(\Rightarrow9⋮x-4\)hay \(x-4\inƯ\left(9\right)\)

Ta có \(x-4\inƯ\left(9\right)\in\left\{\pm1;\pm3;\pm9\right\}\)

\(\Rightarrow x\in\left\{5;3;7;1;13;-5\right\}\)

b, Đặt \(B=\frac{9}{x-4}\)\(\Rightarrow A_{max}\)khi \(B_{max}\)

Vì \(9>0\)để B đặt GTLN \(\Rightarrow\hept{\begin{cases}x-4>0\\\left(x-4\right)_{min}\end{cases}}\)

Mà \(x\in N\)\(\Rightarrow x-4=1\)

\(\Rightarrow x=5\)

\(\Rightarrow B_{max}=\frac{9}{5-4}=9\)

\(\Rightarrow A_{max}=1+9=10\)khi \(x=5\)

c, Đặt \(B=\frac{9}{x-4}\)\(\Rightarrow A_{min}\)khi \(B_{min}\)

Vì \(9>0\)để B đạt GTNN \(\Rightarrow\hept{\begin{cases}x-4< 0\\\left(x-4\right)_{max}\end{cases}}\)

Mà \(x\in N\)\(\Rightarrow x-4\in Z\)

\(\Rightarrow x-4=-1\)

\(\Rightarrow x=3\)

\(\Rightarrow B_{min}=\frac{9}{3-4}=-9\)

\(\Rightarrow A_{min}=1+\left(-9\right)=\left(-8\right)\)khi \(x=3\)

a: Khi x=6 thì \(A=\dfrac{4}{6-3}=\dfrac{4}{3}\)

b: \(B=\dfrac{4x}{x^2-9}-\dfrac{x-3}{x+3}\)(ĐKXĐ: \(x\notin\left\{3;-3\right\}\))

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

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

\(=\dfrac{4x-x^2+6x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{-x^2+10x-9}{\left(x+3\right)\left(x-3\right)}\)

a, để A nguyên

=> 7 - x chia hết cho x - 5

=> 5 - x + 2 chia hết cho x - 5

=> -(x - 5) + 2 chia hết cho x - 5

=> 2 chia hết cho x - 5

=> x - 5 thuộc Ư(2)

=> x - 5 thuộc {-1;1-2;2}

=> x thuộc {4; 6; 3; 7}

\(A=\frac{x+5}{x+2}\left(x\ne-2\right)\)

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

Để A đạt GTLN thì \(\frac{3}{x+2}\)đạt GTLN

=> x+2=3

<=> x=1 (tmđk)
Vậy x=1

=>A=x+2+3/x+2=x+2/x+2+3/x+2=1+3/x+2.

Để Amax=> 3/x+2 max.

=>3/x+2>0.

Mà 3>0=>x+2>0;x+2EZ.(1)

Để 3/x+2 max=>x+2 min.(2)

Từ (1) và (2)=>x+2=1=>x=-1.

=>A=4.

Vậy A max=4 tại x=-1.

k nha có j kb vs mk nha

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 \(-2⋮\sqrt{x}-3\)

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

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

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