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M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)
=\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)
\(M=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}.\left(\frac{1}{1-\sqrt{x}}-1\right)\)
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(M=\frac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(M=\frac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(M=\frac{3\left(x+\sqrt{x}-2\right)}{x+\sqrt{x}-2}\)
\(M=3\)
\(a,ĐK:9x^2-1\ne0\Leftrightarrow x^2\ne\frac{1}{9}\Leftrightarrow x\ne\pm\frac{1}{3}\)
\(b,M=\frac{\sqrt{9x^2-6x+1}}{9x^2-1}=\frac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}\)
với \(3x-1>0\) ta có \(M=\frac{3x-1}{\left(3x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)
với \(3x-1< 0\) ta có \(M=\frac{-\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}=-\frac{1}{3x+1}\)
\(c,\) th1 : \(M=\frac{1}{3x+1}\) khi \(x>\frac{1}{3}\) mà \(M=\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3x+1}=\frac{1}{4}\Leftrightarrow x=1\left(thoaman\right)\)
th2 : \(M=-\frac{1}{3x+1}\) khi \(x< \frac{1}{3}\) mà \(M=\frac{1}{4}\)
\(\Leftrightarrow\frac{-1}{3x+1}=\frac{1}{4}\Leftrightarrow3x+1=-4\Leftrightarrow x=-\frac{5}{3}\left(thoaman\right)\)
\(d,M=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}< 0\) có \(\left|3x-1\right|>0\)
\(\Rightarrow\left(3x-1\right)\left(3x+1\right)< 0\)
th1 : \(\hept{\begin{cases}3x-1>0\\3x+1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{1}{3}\\x< -\frac{1}{3}\end{cases}\left(voli\right)}}\)
th2 : \(\hept{\begin{cases}3x-1< 0\\3x+1>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{1}{3}\\x>-\frac{1}{3}\end{cases}\Leftrightarrow-\frac{1}{3}< x< \frac{1}{3}}\)
a)ĐKXĐ:x khác 4, x>0
\(Q=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right):\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2x}{\left(x-4\right)\left(\sqrt{x}-2\right)}\)
mình nghĩ đề sai nên không làm tiếp nữa
a) ĐKXĐ: x \(\ge\)0; x \(\ne\)4
Ta có: P = \(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\frac{x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-\left(x+6\sqrt{x}+\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}\)
b) Với x \(\ge\)0 và x \(\ne\)4, ta có:
P > -1 <=> \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}>-1\)
<=> \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}+1>0\)
<=> \(\frac{\sqrt{x}-2-\sqrt{x}-6}{\sqrt{x}-2}>0\)
<=> \(\frac{-8}{\sqrt{x}-2}>0\)
Do -8 < 0 => \(\sqrt{x}-2< 0\) <=> \(\sqrt{x}< 2\)<=> \(x< 4\)
mà x \(\ge0\) => 0 \(\le\)x \(< \)4
c)Với x \(\ge\)0 và x \(\ne\)4
Để P \(\in\)Z <=> -8 \(-8⋮\sqrt{x}-2\)
<=> \(\sqrt{x}-2\inƯ\left(-8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Do \(\sqrt{x}\ge0\) <=> \(\sqrt{x}-2\ge-2\) => \(\sqrt{x}-2\in\left\{-2;-1;1;2;4;8\right\}\)
Lập bảng:
| \(\sqrt{x}-2\) | -2 | -1 | 1 | 2 | 4 | 8 |
| x | 0 | 1 | 9 | 16 | 36 | 100 |
Vậy ....
a/ Đkxđ: x\(\ge\)0 x\(\ne\)4
=\(\frac{3\left(\sqrt{x}+2\right)+2\left(\sqrt{x}-2\right)+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{5\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{5}{\sqrt{x}-2}\)
b/ Với x\(\ge\)0 vã\(\ne\)4
Để M\(\in\)Z \(\Leftrightarrow\) \(\frac{5}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\) \(\sqrt{x}-2\inƯ\left(5\right)\)
\(\begin{cases}\sqrt{x}-2=5\\\sqrt{x}-2=-5\\\sqrt{x}-2=1\\\sqrt{x}-2=-1\end{cases}\Rightarrow\begin{cases}x=49\left(tmĐKXĐ\right)\\KhongcogiatriTm\\x=9\left(tmĐKXĐ\right)\\x=1\left(tmĐKXĐ\right)\end{cases}\)
Vậy để M\(\in\)Z thì x=.....
c/ Với...
Để M<2 thì \(\frac{5}{\sqrt{x}-2}< 2\Rightarrow\frac{5-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}< 0\)
\(\left[\begin{array}{nghiempt}\hept{\begin{cases}9-2\sqrt{x}>0\\\sqrt{x}-2< 0\end{array}\right.\\\hept{\begin{cases}9-2\sqrt{x}< 0\\\sqrt{x}-2>0\end{array}\right.\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}\hept{\begin{cases}x< \frac{81}{4}\\x< 4\end{array}\right.\\\hept{\begin{cases}x>\frac{81}{4}\\x>4\end{array}\right.\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x< 4\\x>\frac{81}{4}\end{array}\right.}\)
thanks