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=\(\dfrac{9}{4}\)

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. \(H=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=5\Leftrightarrow\dfrac{\sqrt{a}-2-2}{\sqrt{a}-2}=5\Leftrightarrow1-\dfrac{2}{\sqrt{a}-2}=5\)

\(\Leftrightarrow\dfrac{2}{\sqrt{a}-2}=-4\Leftrightarrow-4\sqrt{a}+8=2\Leftrightarrow-4\sqrt{a}=-6\Leftrightarrow\sqrt{a}=\dfrac{3}{2}\Leftrightarrow a=\dfrac{9}{4}\)

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)

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

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 : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)

(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: \(x=0.\)

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 \(P< 1\)

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 :)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

__

Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)