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bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b) Ta có:
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)
\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)+2\sqrt{x}\cdot\left(\sqrt{x}-3\right)-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{x+5\sqrt{x}+6+2x-6\sqrt{x}-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{7\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7}{\sqrt{x}+3}\)
c) Nếu x không là số chính phương => P vô tỉ (loại)
=> x là số chính phương khi đó để P nguyên thì:
\(\left(\sqrt{x}+3\right)\inƯ\left(7\right)\) , mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\)
\(\Rightarrow\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\)
Vậy x = 16 thì P nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\) ĐKXĐ : x > 0 , x khác 9
\(A=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(A=\frac{-3\sqrt{x}}{\sqrt{x}+3}.\frac{1}{\sqrt{x}+1}\)
\(A=\frac{-3\sqrt{x}}{x+4\sqrt{x}+4}\)
\(A=\frac{-3\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
a) ĐKXĐ : x>hoặc = 0 ; x khác 9
Còn câu b,c,d để vài bữa mình làm tiếp cho bây giờ mình đi ngủ đã buồn ngủ quá !
----------------- -Học tốt-----------------
![](https://rs.olm.vn/images/avt/0.png?1311)
a) đkxđ : \(x\ge0;x\ne2;x\ne1\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{-2x+\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{\left(-2\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
b) P>=2
\(\frac{-2x+\sqrt{x}+3-2\left(x-3\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\ge0\)
\(\frac{-2x+\sqrt{x}+3-2x+6\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\ge0\)
\(\frac{-4x+7\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\ge0\)
\(\frac{-4\left(\sqrt{x}-\frac{7+\sqrt{33}}{8}\right)\left(\sqrt{x}-\frac{7-\sqrt{33}}{8}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\ge0\)
a) Ta có :\(x-3\sqrt{x}+2=\left(\sqrt{x}\right)^2-\sqrt{x}-2\sqrt{x}+2\)\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
P xác định \(\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\\\sqrt{x}-1\ne0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne2\\\sqrt{x}\ne1\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\\x\ne1\end{cases}}}\)
Vậy với \(x\ge0;x\ne4;x\ne1\)thì P xác định
b) Cho mình hỏi, câu b là yêu cầu tìm x để \(P\ge2\)hay chứng minh \(P\ge2\)
c) \(P=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}-\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{x-\sqrt{x}-3\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{\sqrt{x}-2x+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(3-2\sqrt{x}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
Bạn thử xem lại đề nhé. Nếu rút gọn thì kết quả như trên, không rút gọn đc nữa. Chỉ khi nào trên tử là số mới tìm P nguyên đc
Mình sẽ suy nghĩ thêm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)
__
Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)
\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)
Đạt được khi x = 9
2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)
Không có GTLN nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)