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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) \(Q=\) \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne1\right)\)
\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{2}{x-1}\) \(\left(đpcm\right)\).
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:
| x -1 | 1 | -1 | 2 | -2 |
| x | 2(TM) | 0(ko TM) | 3(TM) | -1(koTM) |
Vậy để biểu thức Q nhận giá trị nguyên thì \(x\in\left\{2;3\right\}\)
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
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
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b) \(x=9\Rightarrow A=\dfrac{3}{3+1}=\dfrac{3}{4}\)
\(x=7-4\sqrt{3}\Rightarrow A=\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}+1}=\dfrac{\sqrt{7-2\sqrt{12}}}{\sqrt{7-2\sqrt{12}}+1}=\dfrac{\sqrt{4-2\sqrt{3}\sqrt{4}+3}}{\sqrt{4-2\sqrt{3}\sqrt{4}+3}+1}=\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}=\dfrac{2-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\dfrac{3-\sqrt{3}}{6}\)
Bài 1: \(H=4\sqrt{x}-x-y+6\sqrt{y}-15=-\left(x-4\sqrt{x}+\text{4 }\right)+4-\left(y-6\sqrt{y}+9\right)+9-15=-\left(\sqrt{x}-2\right)^2-\left(\sqrt{y}-3\right)^2-2\le-2\)
Vậy H đạt gtln bằng -2 \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=9\end{matrix}\right.\)
Bài 2:
(+) \(F=\dfrac{2\sqrt{x}-5}{\sqrt{x}-4}=2-\dfrac{1}{\sqrt{x}-4}\)
\(F\in Z\Leftrightarrow\dfrac{1}{\sqrt{x}-4}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-4=-1\\\sqrt{x}-4=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(N\right)\\x=25\left(N\right)\end{matrix}\right.\)
Kl: x=9, x=25
(+) \(G=\dfrac{4\sqrt{x}-9}{\sqrt{x}+4}=\dfrac{4\left(\sqrt{x}+4\right)-16-9}{\sqrt{x}+4}=4-\dfrac{25}{\sqrt{x}+4}\)
\(G\in Z\Leftrightarrow\dfrac{37}{\sqrt{x}+4}\in Z\Leftrightarrow\) (tự làm tiếp nhé)
a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1-2}{\sqrt{3}+1+2}=\dfrac{-3+2\sqrt{3}}{3}\)
c: Q=3
=>3căn x+6=căn x-2
=>2căn x=-8(loại)
d: Q>1/2
=>Q-1/2>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)
=>2căn x-4-căn x-2>0
=>căn x>6
=>x>36
d: Q nguyên
=>căn x+2-4 chia hết cho căn x+2
=>căn x+2 thuộc Ư(-4)
=>căn x+2 thuộc {2;4}
=>x=0 hoặc x=4(nhận)
\(Q=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\left(đk:x\ge0,x\ne4\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{8}{\sqrt{x}-2}=1+\dfrac{8}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
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)