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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
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\(\in Z\)
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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)
\(Q=\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\left(đk:x\ge0,x\ne4\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{8}{\sqrt{x}-2}=1+\dfrac{8}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
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)