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\(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(Q=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{2\sqrt{x}-9-\left(x-9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{2\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
a, Với \(x\ge0;x\ne1\)
\(B=\frac{1}{\sqrt{x}-1}=2\Rightarrow2\sqrt{x}-2=1\Leftrightarrow2\sqrt{x}-3=0\Leftrightarrow x=\frac{9}{4}\)
b, Ta có : \(A.B=\frac{x+3}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=\frac{x+3}{x-1}=\frac{x-1+4}{x-1}=1+\frac{4}{x-1}\)
\(\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x - 1 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 2 | 0 | 3 | -1 | 5 | -3 |
c, Ta có : \(A=\frac{x+3}{\sqrt{x}+1}\le3\Leftrightarrow\frac{x+3}{\sqrt{x}+1}-3\le0\)
\(\Leftrightarrow\frac{x-3\sqrt{x}}{\sqrt{x}+1}\le0\Rightarrow\sqrt{x}-3\le0\Leftrightarrow x\le9\)
Kết hợp với đk vậy 0 =< x =< 9
a) ĐK : x ≥ 0 ; x ≠ 2 ; x ≠ 3
A= \(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)
=\(\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\text{}\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{-1}{\sqrt{x}-3}\)
Vậy...
b)Ta có A<-1
⇒\(\frac{-1}{\sqrt{x}-3}\) <-1
⇒\(\frac{-1}{\sqrt{x}-3}\) +1<0
⇒\(\frac{\sqrt{x}-4}{\sqrt{x}-3}\) <0
⇒\(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\)
⇒9< x <16
Vậy...
c) Ta có A = \(\frac{-1}{\sqrt{x}-3}\)
⇒2A=\(\frac{-2}{\sqrt{x}-3}\)
Để 2A ∈ Z thì \(\frac{-2}{\sqrt{x}-3}\) ∈ Z
⇒\(\sqrt{x}-3\) ∈ Ư(-2) =\(\left\{1;-1;2;-2\right\}\)
Ta có bảng
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 |
| x | 16(tm) | 4(tm) | 25(tm) | 1(tm) |
Vậy...
OK!!!
đó bạn