Rút gọn P
P = \(\left(\dfrac{a+3\sqrt{a}+2}{a+\sqrt{a}-2}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
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16 tháng 9 2021
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
Để A nguyên
\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)
Do \(\sqrt{a}>0,\sqrt{a}\ne1\)
\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)
\(\Leftrightarrow a=4\)
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: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
4 tháng 9 2021
a: Ta có: \(P=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{a+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{\left(a+1\right)\cdot\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
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: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
15 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
15 tháng 8 2021
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
15 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
4 tháng 10 2021
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
4 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2022
ĐKXĐ: \(-1\le a< 1\); \(a\ne0\)
\(P=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}^2}{\sqrt{\left(1-a\right)\left(1+a\right)}-\sqrt{1-a}^2}\right).\left(\sqrt{\dfrac{\left(1-a\right)\left(1+a\right)}{a^2}}-\dfrac{1}{a}\right)\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\sqrt{\dfrac{\left(1-a\right)\left(1+a\right)}{a^2}}-\dfrac{1}{a}\right)\)
\(=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)
\(=\dfrac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{1+a-\left(1-a\right)}.\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)
\(=\left(\dfrac{1+\sqrt{\left(1-a\right)\left(1+a\right)}}{a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)
- Với \(a>0\)
\(\Rightarrow P=\dfrac{\left(\sqrt{1-a^2}+1\right)\left(\sqrt{1-a^2}-1\right)}{a^2}=\dfrac{1-a^2-1}{a^2}=-1\)
- Với \(a< 0\)
\(\Rightarrow P=-\dfrac{\left(1+\sqrt{1-a^2}\right)^2}{a^2}\)
Với \(a>0;a\ne1\):
\(P=\left(\dfrac{a+3\sqrt{a}+2}{a+\sqrt{a}-2}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\\ =\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]:\left[\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\\ =\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)