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Câu hỏi của Trần Nam Hải - Toán lớp 9 - Học toán với OnlineMath
https://olm.vn/hoi-dap/detail/228029923283.html
- Chúc bạn học tốt -
_Minh ngụy_
\(A=\frac{1}{1-\sqrt{a}}+\frac{a\sqrt{a}}{\sqrt{a}-1}\left(\text{ĐKXĐ: a}\ne1\right)\)
\(A=\frac{-1}{-\left(1-\sqrt{a}\right)}+\frac{\left(\sqrt{a}\right)^2.\sqrt{a}}{\sqrt{a}-1}\)
\(A=\frac{-1}{\sqrt{a}-1}+\frac{\sqrt{a}^3}{\sqrt{a}-1}\)
\(A=\frac{-1+\sqrt{a}^3}{\sqrt{a}-1}\)
\(A=\frac{\sqrt{a}^3-1}{\sqrt{a}-1}\)
\(A=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}\)
\(A=a+\sqrt{a}+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\right):\left(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\right)\left(1\right)\)
a) B xác định khi và chỉ khi :
\(\left\{{}\begin{matrix}a\ge0\\\sqrt[]{a}\ne0\\\sqrt[]{a}-1\ne0\\\sqrt[]{a}-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
b) \(\left(1\right)\Leftrightarrow B=\left(\dfrac{\sqrt[]{a}-\left(\sqrt[]{a}-1\right)}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{\left(\sqrt[]{a}+1\right)\left(\sqrt[]{a}-1\right)-\left(\sqrt[]{a}+2\right)\left(\sqrt[]{a}-2\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{a-1-\left(a-4\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{3}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right).\left(\dfrac{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}{3}\right)\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện : a> 0 ; a khác 1
\(A=\frac{\left(\sqrt{a}\right)^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}\right)^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{2a+2}{a-1}\right)\)
\(A=\frac{2\sqrt{a}}{\sqrt{a}}+\frac{2\left(a+1\right)}{\sqrt{a}}=2+\frac{2\sqrt{a}\left(a+1\right)}{a}\)
\(A=\frac{1}{1-\sqrt{a}}+\frac{a\sqrt{a}}{\sqrt{a}-1}\left(a\ge0,a\ne1\right)\)
\(A=\frac{-1}{\sqrt{a}-1}+\frac{a\sqrt{a}}{\sqrt{a}-1}\)
\(A=\frac{-1+a\sqrt{a}}{\sqrt{a}-1}=\frac{a\sqrt{a}-1}{\sqrt{a}-1}\)
\(A=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}\)
\(A=\frac{a+\sqrt{a}+1}{1}=a+\sqrt{a}+1\)
Vậy.............
-Chúc bạn học tốt >.<-
_Minh ngụy_