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a) đk: \(a>0;a\ne1\)
b) Xét K = \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
= \(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}.\left(\sqrt{a}-1\right)\)
= \(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
Xét \(a=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2\)
<=> \(\sqrt{a}=1+\sqrt{2}\)
<=> K = \(\dfrac{\left(\sqrt{2}+2\right)\sqrt{2}}{\sqrt{2}+1}=2\)
c) Đẻ K < 0
<=> \(\dfrac{a-1}{\sqrt{a}}< 0\)
Mà \(\sqrt{a}>0\)
<=> a < 1
<=> 0 < a < 1
a: \(K=\dfrac{a-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1+2}{a-1}\)
\(=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}+3}\)
\(=\dfrac{a\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
c: Vì \(\sqrt{a}+3>=3>0;\sqrt{a}>0;a\sqrt{a}+1>0\)
nên K>0 với mọi a thỏa mãn ĐKXĐ
=>Không có giá trị nào của a để K<0
Lời giải:
a) ĐK: \(a>0; a\neq 1\)
\(K=\left(\frac{a}{\sqrt{a}(\sqrt{a}-1)}-\frac{1}{\sqrt{a}(\sqrt{a}-1)}\right): \left(\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}+\frac{2}{(\sqrt{a}-1)(\sqrt{a}+1)}\right)\)
\(=\frac{a-1}{\sqrt{a}(\sqrt{a}-1)}: \frac{\sqrt{a}+1+2}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(=\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}(\sqrt{a}-1)}. \frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}+3}\)
\(=\frac{(\sqrt{a}+1)^2(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}+3)}\)
b) \(a=3+2\sqrt{a}\Leftrightarrow a-2\sqrt{a}-3=0\)
\(\Leftrightarrow (\sqrt{a}-3)(\sqrt{a}+1)=0\)
\(\Rightarrow \sqrt{a}=3\)
Khi đó: \(K=\frac{(3+1)^2(3-1)}{3.(3+3)}=\frac{16}{9}\)
c) Để \(K< 0\Leftrightarrow \frac{(\sqrt{a}+1)^2(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}+3)}< 0\)
Mà \(\frac{(\sqrt{a}+1)^2}{\sqrt{a}(\sqrt{a}+3)}>0, \forall a> 0; a\neq 1\), do đó \(\sqrt{a}-1< 0\Leftrightarrow 0< a< 1\)
Vậy .........
Lời giải:
ĐK: \(a>0; a\neq 1\)
a) \(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right): \left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(B=\left(\frac{a}{a-\sqrt{a}}-\frac{1}{a-\sqrt{a}}\right): \left(\frac{\sqrt{a}-1}{(\sqrt{a}+1)(\sqrt{a}-1)}+\frac{2}{a-1}\right)\)
\(=\frac{a-1}{a-\sqrt{a}}:\left(\frac{\sqrt{a}-1}{a-1}+\frac{2}{a-1}\right)\)
\(=\frac{a-1}{a-\sqrt{a}}: \frac{\sqrt{a}+1}{a-1}=\frac{a-1}{a-\sqrt{a}}.\frac{a-1}{\sqrt{a}+1}=\frac{(a-1)^2}{\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}=\frac{(a-1)^2}{\sqrt{a}(a-1)}=\frac{a-1}{\sqrt{a}}\)
b) Ta có:
\(a=3+2\sqrt{2}=2+1+2\sqrt{2}=(\sqrt{2}+1)^2\)
\(\Rightarrow K=\frac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\frac{2+2\sqrt{2}}{\sqrt{2}+1}=\frac{2(1+\sqrt{2})}{\sqrt{2}+1}=2\)
c) \(K< 0\leftrightarrow \frac{a-1}{\sqrt{a}}< 0\Leftrightarrow a-1< 0\) (do \(\sqrt{a}>0\))
\(\Leftrightarrow a< 1\)
Vậy \(0< a< 1\)
Nhật Hạ : bạn ghi trên đề bài mà.
Thực ra nó chỉ là tên biểu thức nên không quan trọng.
a: \(K=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{a-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\)
b: Thay \(a=3+2\sqrt{2}\) vào K, ta được:
\(K=\dfrac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
c: Để K<0 thì a-1<0
hay 0<a<1
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
a) \(K=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1}{a-1}-\dfrac{2}{a-1}\right)=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{a-1}{\sqrt{a}-3}=\dfrac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
b) Ta có: \(\sqrt{a}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)
Thay \(a=3+2\sqrt{2}\) và \(\sqrt{a}=\sqrt{2}+1\) vào K:
\(K=\dfrac{\left(\sqrt{2}+1+1\right)\left(3+2\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1-3\right)}=\dfrac{\left(\sqrt{2}+2\right)\left(2\sqrt{2}+2\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-2\right)}=\dfrac{2\left(\sqrt{2}+2\right)}{\sqrt{2}-2}=\dfrac{2\left(1+\sqrt{2}\right)}{1-\sqrt{2}}\)
c) Đk: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\\a\ne9\end{matrix}\right.\)
Vì \(\left\{{}\begin{matrix}\sqrt{a}+1>0\\\sqrt{a}>0\end{matrix}\right.\)
Nên, để K<0 thì \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1>0\\\sqrt{a}-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1< 0\\\sqrt{a}-3>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>1\\\sqrt{a}< 3\end{matrix}\right.\\\left\{{}\begin{matrix}a< 1\\\sqrt{a}>3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>1\\a< 9\end{matrix}\right.\\\left\{{}\begin{matrix}a< 1\\a>9\end{matrix}\right.\left(vn\right)}\end{matrix}\right.\)\(\Leftrightarrow1< a< 9\)
Kl: \(1< a< 9\)
chỗ công thức bị lỗi bạn tự giải nhé (dễ mà ^^! chỗ căn bình phương 2 vế lên thôi ), nãy giờ cứ sửa đi sửa lại mệt quá T_T!!