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![](https://rs.olm.vn/images/avt/0.png?1311)
a,Đk: a≥0 ; a khác 4
H=\(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\) -\(\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\) -\(\dfrac{1}{\sqrt{a}-2}\)
= \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
=\(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
=\(\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
=\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b, Để H<2
<=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) <2
<=> \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) -2<0
<=>\(\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}\) <0
<=>\(\dfrac{-\sqrt{a}}{\sqrt{a}-2}\) <0
<=>\(\left\{{}\begin{matrix}-\sqrt{a}< 0\\\sqrt{a}-2>0\end{matrix}\right.\) ( vì \(\sqrt{a}>0< =>-\sqrt{a}< 0\)
<=> a>4
vậy để H <2 khi a>4
c, Ta có a\(^2\) +3a=0
<=> a(a+3)=0
<=>a=0 hoặc a=-3(vô lí)
+ Với a=0 <=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) =\(\dfrac{0-4}{0-2}\) =2
d, Để H=5
<=> \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) =5
<=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) -5=0
<=>\(\dfrac{\sqrt{a}-4-5\sqrt{a}+10}{\sqrt{a}-2}\) =0
<=>-4\(\sqrt{a}\) +6=0
<=> a=\(\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(ĐKXĐ:a\ge0,a\ne2\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-4-8-\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\left(a-4\sqrt{a}\right)+\left(3\sqrt{a}-12\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\sqrt{a}\left(\sqrt{a}-4\right)+3\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b. Mk nghĩ là H < 2 chứ
\(H=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}< 2\)
\(\Leftrightarrow\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}=\dfrac{-\sqrt{a}}{\sqrt{a}-2}< 0\)
\(\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
c. \(a^2+3a=0\Leftrightarrow a\left(a+3\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\left(n\right)\\a=-3\left(l\right)\end{matrix}\right.\)
Thay \(a=0\) và H ta được:
\(\dfrac{0-4}{0-2}=2\)
d. \(H=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=5\Leftrightarrow\dfrac{\sqrt{a}-2-2}{\sqrt{a}-2}=5\Leftrightarrow1-\dfrac{2}{\sqrt{a}-2}=5\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}-2}=-4\Leftrightarrow-4\sqrt{a}+8=2\Leftrightarrow-4\sqrt{a}=-6\Leftrightarrow\sqrt{a}=\dfrac{3}{2}\Leftrightarrow a=\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{-\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt{2}}\)
\(=\dfrac{-2}{\sqrt{a}+1}\)
b) Ta có: \(P=\dfrac{-1}{2}\)
nên \(\dfrac{2}{\sqrt{a}+1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{a}+1=4\)
\(\Leftrightarrow a=9\)(thỏa ĐK)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để H<2 thì H-2<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}< 0\)
=>căn a-2>0
hay a>4
d: Để H=5 thì căn a-4=5 căn a-10
=>-4 căn a=-6
=>căn a=3/2
hay a=9/4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Rut gon H
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)
Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x}+3}.\)
(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).
Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)
\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)
(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)
\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).
Vậy: \(x\in\varnothing.\)
(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))
\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)
(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)
Vậy: \(x=0.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(a\ge0;a\ne4\)
a) ⇔ \(P=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
⇔ \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b) \(P< 1\Leftrightarrow\dfrac{\sqrt{a}-4}{\sqrt{a}-2}< 1\)
\(\Leftrightarrow\dfrac{\sqrt{a}-4}{\sqrt{a}-2}-1< 0\Leftrightarrow\dfrac{-2}{\sqrt{a}-2}< 0\)
Do \(-2< 0\) ⇔ \(\sqrt{a}-2< 0\Leftrightarrow a< 4\)
Kết hợp điều kiện ban đầu, ta có: \(0< a< 4\)
Vậy khi \(0< a< 4\) thì \(P< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chắc đề em gõ bị lỗi nhỏ :) Cô sẽ sửa nhé :)
a. ĐK: \(a\ge0,a\ne4\)
\(H=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{a+\sqrt{a}-6}=\frac{a-4-4-\sqrt{a}-3}{a+\sqrt{a}-6}\)
\(=\frac{a-\sqrt{a}-12}{a+\sqrt{a}-6}=\frac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)
b. \(H< 2\Leftrightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}< 2\Leftrightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}-2< 0\Leftrightarrow\frac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}< 0\)
\(\Leftrightarrow\frac{-\sqrt{a}}{\sqrt{a}-2}< 0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow x>4\)
Tương tự với các câu còn lại nhé :)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)
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Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)
\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)