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ĐKXĐ: \(x\ge0;a\ne4\)
\(H=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)
\(H< 2\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}< 2\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}-2< 0\)
\(\Rightarrow\frac{\sqrt{a}-4-2\left(\sqrt{a}-2\right)}{\sqrt{a}-2}< 0\Rightarrow\frac{-\sqrt{a}}{\sqrt{a}-2}< 0\)
\(\Rightarrow\frac{\sqrt{a}}{\sqrt{a}-2}>0\Rightarrow\sqrt{a}-2>0\Rightarrow a>4\)
\(a^2+3a=0\Rightarrow a\left(a+3\right)=0\Rightarrow a=0\) (do \(a\ge0\Rightarrow a+3>0\))
\(\Rightarrow H=\frac{0-4}{0-2}=2\)
\(H=5\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}=5\Rightarrow\sqrt{a}-4=5\sqrt{a}-10\)
\(\Rightarrow4\sqrt{a}=6\Rightarrow\sqrt{a}=\frac{3}{2}\Rightarrow a=\frac{9}{4}\)
a. ĐKXĐ : \(a\ge0;a\ne1;a\ne4;a\ne9\)
\(C=-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}-\frac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)}{\sqrt{a}-2}+\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\)
\(=-\sqrt{a}-1-a-2\sqrt{a}-4+\sqrt{a}-2\)
\(=-a-2\sqrt{a}-7\)
b. \(C=-a-2\sqrt{a}-7\le-7\)
\(\text{Dấu }=\text{xảy ra }\Leftrightarrow a=0\)
c. \(a^2-3a=0\)
\(\Leftrightarrow a\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}C=-7\\C=-10-2\sqrt{3}\end{matrix}\right.\)
d. \(C=-13\)
\(\Leftrightarrow-a-2\sqrt{a}-7=-13\)
\(\Leftrightarrow a+2\sqrt{a}-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-1+\sqrt{7}\left(TM\right)\\a=-1-\sqrt{7}\left(\text{loại}\right)\end{matrix}\right.\)
\(\Leftrightarrow a=-1+\sqrt{7}\)
\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
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. \(H=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=5\Leftrightarrow\dfrac{\sqrt{a}-2-2}{\sqrt{a}-2}=5\Leftrightarrow1-\dfrac{2}{\sqrt{a}-2}=5\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}-2}=-4\Leftrightarrow-4\sqrt{a}+8=2\Leftrightarrow-4\sqrt{a}=-6\Leftrightarrow\sqrt{a}=\dfrac{3}{2}\Leftrightarrow a=\dfrac{9}{4}\)
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 :)