\(Q=\frac{a-1}{a-3\sqrt{a}}\)
tìm a để Q >1
tìn a để Q<1
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\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
1)đặt nhân tử chung quy đồng là xong
2)phân tích x+2cănx-3=(1-cănx)(3+cănx)
3)2a+căn a đặt căn a ra r rút gọn
\(P=\left(\frac{\sqrt{a}}{3+\sqrt{a}}+\frac{a+9}{9-a}\right):\left(\frac{3\sqrt{a}+1}{a-3\sqrt{a}}-\frac{1}{\sqrt{a}}\right)\)
\(P=\left[\frac{\sqrt{a}\left(3-\sqrt{a}\right)}{9-a}+\frac{a+9}{9-a}\right]:\left[\frac{3\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-3\right)}-\frac{1}{\sqrt{a}}\right]\)
\(P=\frac{3\sqrt{a}-a+a+9}{9-a}:\left[\frac{3\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}\right]\)
\(P=\frac{3\sqrt{a}+9}{9-a}:\frac{2\sqrt{a}+4}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
\(P=\frac{3\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(3-\sqrt{a}\right)}.\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{2\sqrt{a}+4}\)
\(P=\frac{-3\sqrt{a}}{2\sqrt{a}+4}\)
b) theo câu a) \(P=\frac{-3\sqrt{a}}{2\sqrt{a}+4}\) với \(ĐKXĐ:a\ge0;a\ne9\)
theo bài ra \(P< -1\Leftrightarrow\frac{-3\sqrt{a}}{2\sqrt{a}+4}< -1\)
\(\Rightarrow\frac{-3\sqrt{a}}{2\sqrt{a}+4}+1< 0\)
\(\Rightarrow\frac{-3\sqrt{a}+2\sqrt{a}+4}{2\sqrt{a}+4}< 0\)
\(\Rightarrow\frac{4-\sqrt{a}}{2\sqrt{a}+4}< 0\)
\(\Rightarrow4-\sqrt{a}< 0\)
\(\Rightarrow-\sqrt{a}< -4\)
\(\Rightarrow\sqrt{a}>4\)
\(\Rightarrow a>16\)
kết hợp đkxđ a>0 và a\(\ne9\)ta có:
với \(a>16\) \(\Leftrightarrow\hept{\begin{cases}0< a< 16\\a\ne9\end{cases}}\)
a, Với x > 0
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1}{x+\sqrt{x}}=\frac{x-1+1}{x+\sqrt{x}}=\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
b, Ta có : \(A>\frac{2}{3}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{2}{3}>0\Leftrightarrow\frac{3\sqrt{x}-2\sqrt{x}-2}{3\left(\sqrt{x}+1\right)}>0\)
\(\Rightarrow\sqrt{x}-2>0\Leftrightarrow x>4\)
c, \(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}+3}{2\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}+2}=\frac{2\sqrt{x}+6}{2\sqrt{x}+2}=1+\frac{4}{2\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+1}\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\sqrt{x}+1\) | 1 | 2 |
\(\sqrt{x}\) | 0 (loại ) | 1 |
x | loại | 1 |
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b. \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
c. Để \(Q< 1\Rightarrow Q-1< 0\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Vậy \(0\le x< 9\)thì \(Q< 1\)