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\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
Không thể tìm được GTLN hay GTNN của Q.
b)
\(\frac{3x+3}{\sqrt{x}}=3\sqrt{x}+\frac{3}{\sqrt{x}}\)
Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)
Vậy x=1, x=9 là các giá trị cần tìm
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Để biểu thức đã cho nhận giá trị nguyên buộc \(\dfrac{4}{\sqrt{x}-3}\) nguyên
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{1;4;16;25;49\right\}\)
Vậy ......
Ta có
\(1D=\frac{\sqrt{x}-2}{\sqrt{x}-3}=1+\frac{1}{\sqrt{x}-3}\)
Để cho D nguyên thì \(\sqrt{x}-3\)phải là ước của 1
\(\Rightarrow\sqrt{x}-3=\left(-1;1\right)\)
=> x = (4; 16)
=> D = (0; 2)
1/ Để N nhận giá trị nguyên thì trước hết \(\sqrt{x}-2\)phải là ước của 3
\(\sqrt{x}-2=\left(-3;-1;1;3\right)\)
Thế vào ta tìm được x = (1; 9; 25)
=> N = (- 3; 3;1)
a,\(P=\frac{7}{\sqrt{x}+3}\Rightarrow\sqrt{x}+3\inƯ\left(7\right)=\left\{1;7\right\}\)
| \(\sqrt{x}+3\) | 1 | 7 |
| x | loại | 16 |
b, Ta có : \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3>0\Rightarrow\hept{\begin{cases}\frac{7}{\sqrt{x}+3}\le\frac{7}{3}\\\frac{7}{\sqrt{x}+3}>0\end{cases}}\)
\(\Rightarrow0< P\le\frac{7}{3}\)mà \(P\in Z\)=> \(P\in\left\{1;2\right\}\)
Với \(P=\frac{7}{\sqrt{x}+3}=1\Rightarrow7=\sqrt{x}+3\Leftrightarrow x=16\)( tm )
Với \(P=\frac{7}{\sqrt{x}+3}=2\Rightarrow7=2\sqrt{x}+6\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)( ktm )
\(P=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(P=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\inℤ\Leftrightarrow x+4\sqrt{x}+3⋮\sqrt{x}\)
Giải tiếp nhé sau đó thử chọn :V
\(p=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}}=1+\frac{3}{\sqrt{x}}\)
Để \(x\in Z\Rightarrow P\in Z\)
\(\Rightarrow\sqrt{x}\inƯ\left(3\right)= \left\{-3;3\right\}\)
\(\Leftrightarrow x=9\left(t.mĐKXĐ\right)\)
\(P=\left(\frac{3x+3}{x-9}-\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{3-\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right).ĐKXĐ:x\ge0,x\ne9\)
\(=\left(\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\frac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3}{\sqrt{x}+3}\)
\(b,x=20-6\sqrt{11}=11-2.3\sqrt{11}+9\)
\(=\left(\sqrt{11}-3\right)^2\)
\(P=\frac{3}{\sqrt{x}+3}=\frac{3}{\sqrt{\left(\sqrt{11}-3\right)^2}+3}=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)
\(c,P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+3}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)\(\Leftrightarrow\frac{3-\sqrt{x}}{2\left(\sqrt{x}+3\right)}>0\)
vì \(2\left(\sqrt{x}+3\right)>0\) (nếu x=0 =>pt vô nghiệm)
\(\Rightarrow3-\sqrt{x}>0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
Kết hợp ĐKXĐ: \(0< x< 9\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)
Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow2⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-5;-4;-2;-1\right\}\\ \Leftrightarrow x\in\left\{1;4;16;25\right\}\)
Vậy \(x\in\left\{1;4;16;25\right\}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\)
Tick plz
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+3\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne-3\left(loại\right)\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(x\in Z\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}+3\right)\)
\(\Rightarrow\left(\sqrt{x}+3-2\right)⋮\left(\sqrt{x}+3\right)\)
Vì \(\Rightarrow\left(\sqrt{x}+3\right)⋮\left(\sqrt{x}+3\right)\)
\(\Rightarrow2⋮\left(\sqrt{x}+3\right)\Rightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng:
Vậy không có x nguyên thỏa mãn đề bài