Tìm a để H=\(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)<2
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CH
Cô Hoàng Huyền
Admin
VIP
17 tháng 6 2016
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 :)
P
3 tháng 9 2018
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\).\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)
=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)
=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)
=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)
=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{-\left(a-1\right)}{1}\)= - a + 1
hok tốt
28 tháng 12 2015
a, ĐK: \(a\ge0;\) a khác 4
b,\(A=\frac{a+3\sqrt{a}+2+2a-4\sqrt{a}-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}+2\right)}=\frac{3a-6\sqrt{a}}{\left(\sqrt{a}+2\right)\cdot\left(\sqrt{a}-2\right)}=\frac{3\sqrt{a}}{\sqrt{a}+2}\)
c, để A= 2 KHI \(\frac{3\sqrt{a}}{\sqrt{a}+2}=2\)
<=>\(3\sqrt{a}=2\sqrt{a}+4\)
<=>\(\sqrt{a}=4\)
<=>a=16
tick nha
NT
0
1 tháng 8 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(A=\)\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}.\sqrt{a}}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{\left(2\sqrt{a}\right)^2}\left(\frac{a-2\sqrt{a}+1-a-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2.-4\sqrt{a}}{4a\left(a-1\right)}=\frac{a-1}{\sqrt{a}}\)
\(b,A< 0\Rightarrow\frac{a-1}{\sqrt{a}}< 0\)
Mà \(\sqrt{a}\ge0\Rightarrow a-1\le0\Rightarrow a\le1\)
\(A=2\Rightarrow\frac{a-1}{\sqrt{a}}=2\)
\(\Rightarrow a-1=2\sqrt{a}\Rightarrow a-2\sqrt{a}-1=0\)
\(\Rightarrow a-2\sqrt{a}+1-2=0\)
\(\Rightarrow\left(\sqrt{a}-1\right)^2-\sqrt{2}^2=0\)
\(\Rightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=1+\sqrt{2}\\\sqrt{a}=1-\sqrt{2}\end{cases}\Rightarrow\orbr{\begin{cases}a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\\a=\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}\end{cases}}}\)
BA
1 tháng 8 2019
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{a-1}\)
\(=\frac{a-1}{4a}.\frac{2\sqrt{a}.\left(-2\right)}{1}\)
\(=\frac{a-1}{4a}.\frac{-4\sqrt{a}.}{1}\)
\(=\frac{1-a}{\sqrt{a}}\)
1 tháng 9 2016
a)A=\(\left(\frac{\sqrt{a}^2-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
=\(\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
=\(\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{-4\sqrt{a}}{a-1}\right)\)
=\(\frac{a-1}{\sqrt{a}}\cdot\left(-1\right)\)
=\(\frac{1-a}{\sqrt{a}}\)
1 tháng 9 2016
b) để A<0 thì (ĐKXĐ a#0 a#1
\(\frac{1-a}{\sqrt{a}}< 0\)
mà \(\sqrt{a}>0\)
=> 1-\(\sqrt{a}< 0\)
=> \(\sqrt{a}>1\)
=> a>1
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 10 2019
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 \(a\ge0\Rightarrow a+3>0\))
\(\Rightarrow H=\frac{0-4}{0-2}=2\)
\(H=5\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}=5\Rightarrow\sqrt{a}-4=5\sqrt{a}-10\)
\(\Rightarrow4\sqrt{a}=6\Rightarrow\sqrt{a}=\frac{3}{2}\Rightarrow a=\frac{9}{4}\)
\(\frac{\sqrt{a}-4}{\sqrt{a}-2}< 2\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{\sqrt{a}-2}-\frac{2}{\sqrt{a}-2}< 2\)
\(\Leftrightarrow\)\(1-\frac{2}{\sqrt{a}-2}< 2\)
\(\Leftrightarrow\)\(\frac{2}{\sqrt{a}-2}>1\)
\(\Leftrightarrow\)\(2:\frac{2}{\sqrt{a}-2}< 2:1\)
\(\Leftrightarrow\)\(\sqrt{a}-2< 2\)
\(\Leftrightarrow\)\(\sqrt{a}< 4\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}a< 4^2=16\\a>-4^2=-16\end{cases}}\)
Vậy để \(H< 2\) thì \(a< 16\) hoặc \(a>-16\)
Chúc bạn học tốt ~