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![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(P=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\inℤ\Leftrightarrow x+4\sqrt{x}+3⋮\sqrt{x}\)
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 \(x\in Z\Rightarrow P\in Z\)
\(\Rightarrow\sqrt{x}\inƯ\left(3\right)= \left\{-3;3\right\}\)
\(\Leftrightarrow x=9\left(t.mĐKXĐ\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x>0; x\neq 4$
Sửa lại đề 1 chút.
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{2}{\sqrt{x}+2}\)
\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)
Với mọi $x>0$ thì hiển nhiên $B>0$. Mặt khác, $\sqrt{x}+2\geq 2$ nên $B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}=\frac{7}{3}$
Vậy $0< B\leq \frac{7}{3}$. $B$ đạt giá trị nguyên thì $B=1;2$
$B=1\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=1$
$\Leftrightarrow x=\frac{64}{9}$ (thỏa mãn)
$B=2\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=2$
$\Leftrightarrow x=\frac{1}{9}$ (thỏa mãn)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=2\)
=> Với mọi \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)thì P = 2
Đề sai à --
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đk \(x>0\)và \(x\ne4\)
=\(\left(\frac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right)\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)
=\(\frac{2\sqrt{x}}{x-4}\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)
=\(\frac{2}{\sqrt{x}+2}\)
b) Để \(\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)
\(\Leftrightarrow\frac{4-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)\(>0\)
\(\Leftrightarrow\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)
Vì \(2\left(\sqrt{x}+2\right)>0\)
mà\(\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)
nên \(-\sqrt{x}+2>0\)\(\Leftrightarrow x< 4\)
Vậy vs \(0< x< 4\)thì \(A>\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình nghĩ nên bổ sung x nguyên
\(A=\frac{3\sqrt{x}-2}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)-5}{\sqrt{x}+1}=3-\frac{5}{\sqrt{x}+1}\)
Để A nguyên thì \(\frac{5}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\in\left\{1;5\right\}\Leftrightarrow x=0\)
Thay x = 0 vào thì ta có A âm vậy ............
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)
2) Để \(P=2\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)
\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)
3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)
Thay \(x=\frac{1}{4}\)vào P, ta được :
\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)
4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)
\(\Leftrightarrow9x-3\sqrt{x}-6=0\)
\(\Leftrightarrow3x-\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\)
\(\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow9x^2-13x+4=0\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)
Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)
Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)
5) Để biểu thức nhận giá trị nguyên
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)
\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)
\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)
\(\Leftrightarrow8⋮2-\sqrt{x}\)
\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
Ta loại các giá trị < 0
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)
\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
\(\)
\(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}=\frac{\sqrt{x}-1+4}{\sqrt{x}-1}=1+\frac{4}{\sqrt{x}-1}\)
Để P đạt giá trị nguyên thì \(\frac{4}{\sqrt{x}-1}\) đạt giá trị nguyên
<=>4 chia hết cho \(\sqrt{x}-1\)
<=>\(\sqrt{x}-1\inƯ\left(4\right)\)
<=>\(\sqrt{x}-1\in\left\{-4;-2;-1;1;2;4\right\}\)
<=>\(\sqrt{x}\in\left\{-3;-1;0;2;3;5\right\}\)
<=>\(x\in\left\{0;4;9;25\right\}\)
Cách giải lớp 6 á, thông cảm :)
rút gọn A= ( \(\left(\sqrt{26}+5\sqrt{2}\right)\sqrt{19-5\sqrt{13}}\)