Rút gọn:
\(A=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
K
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Những câu hỏi liên quan
10 tháng 8 2018
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
10 tháng 8 2018
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
21 tháng 9 2021
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) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
5 tháng 5 2021
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
5 tháng 5 2021
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
LD
26 tháng 7 2021
A=\(\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(a-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(a+\sqrt{a}\right)}{\left(a-1\right)}\right]\)::::::::\(\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
=\(\left[\dfrac{1}{\sqrt{a}-1}\right]:\left(\dfrac{2\sqrt{a}}{a-1}\right)\)=\(\dfrac{\sqrt{a}-1}{2\sqrt{a}}\)
=\(\dfrac{a^2+a\sqrt{a}+11a+6}{2\sqrt{a}\left(\sqrt{a}+2\right)}\)
26 tháng 7 2021
Ta có: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
25 tháng 5 2022
\(A=\left[1:\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right)\right]\cdot\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\)
4 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
ND
Nguyễn Đức Trí
VIP
26 tháng 8 2023
\(B=\left(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\right):\left(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\right)\left(1\right)\)
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) \(\left(1\right)\Leftrightarrow B=\left(\dfrac{\sqrt[]{a}-\left(\sqrt[]{a}-1\right)}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{\left(\sqrt[]{a}+1\right)\left(\sqrt[]{a}-1\right)-\left(\sqrt[]{a}+2\right)\left(\sqrt[]{a}-2\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{a-1-\left(a-4\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{3}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right).\left(\dfrac{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}{3}\right)\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}\)
9 tháng 6 2017
\(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\dfrac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=\dfrac{\left(1-a\right)\left(\sqrt{a}+a+1\right)}{1-a}\)
=> \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}=a+2\sqrt{a}+1=\left(\sqrt{a}+1\right)^2\)
Tương tự \(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}=\left(\sqrt{a}-1\right)^2\)
biểu thức trong dấu ngoặc vuông = \(\left[\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)\right]^2=\left(a-1\right)^2\)
\(E=\dfrac{1-a^2}{\left(a-1\right)^2}\)
26 tháng 12 2021
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 12 2021
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)