Rút gọn \(P=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)
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29 tháng 1 2022
\(P=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\left(\sqrt{a}+4\right)}\)
\(=\dfrac{-8\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}\cdot\dfrac{\sqrt{a}+4}{-\left(\sqrt{a}+1\right)}=\dfrac{8}{\sqrt{a}-4}\)
23 tháng 7 2016
Bài 1
a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\) (ĐK : x\(\ge0\) ; x\(\ne\) 1)
\(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)
Mà Ư(2)={-1;1;2;-1}
=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-1\) | 1 | -1 | 2 | -2 |
| a | 4 | 0 | 9 | \(\sqrt{a}=-1\) (ktm) |
vậy a={0;4;9} thì P nguyên
23 tháng 7 2016
Bài 2
\(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)
\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)
\(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)
\(=\frac{2a}{\sqrt{a-4}}\)
30 tháng 6 2016
P= (\(\frac{3\sqrt{a}}{\sqrt{a}+4}+\frac{\sqrt{a}}{\sqrt{a}-4}+\frac{4\left(a+2\right)}{16-a}\)):\(\left(1-\frac{2\sqrt{a}+5}{\sqrt{a}-4}\right)\)
=\(\left(\frac{3\sqrt{a}\left(\sqrt{a}-4\right)}{a-16}+\frac{\sqrt{a}\left(\sqrt{a}+4\right)}{a-16}-\frac{4a+8}{a-16}\right):\left(\frac{\sqrt{a}-4-2\sqrt{a}-5}{\sqrt{a}-4}\right)\)
= \(\left(\frac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{a-16}\right):\left(\frac{-\sqrt{a}-9}{\sqrt{a}-4}\right)\)
=\(\left(\frac{-8\sqrt{a}-8}{a-16}\right).\left(\frac{\sqrt{a}-4}{-\sqrt{a}-9}\right)=\frac{8\sqrt{a}+8}{\left(\sqrt{a}+4\right).\left(\sqrt{a}+9\right)}=\frac{8\sqrt{a}+8}{a+13\sqrt{a}+36}\)
10 tháng 10 2016
Q = \(\frac{\sqrt{a}+3}{\sqrt{a}-2}\)- \(\frac{\sqrt{a}-1}{\sqrt{a}+2}\)+ \(\frac{4-4\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
= \(\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)+4-4\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
=\(\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2+4-4\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
= \(\frac{8+4\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
= \(\frac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\)
= \(\frac{4}{\sqrt{a}-2}\)
26 tháng 5 2017
\(Q=\frac{\sqrt{a+3}}{\sqrt{a-2}}-\frac{\sqrt{a-1}}{\sqrt{a+2}}+\frac{4-4\sqrt{a}}{\left(\sqrt{a-2}\right)\left(\sqrt{a+2}\right)}\)
\(Q=\frac{\left(\sqrt{a+3}\right)\left(\sqrt{a+2}\right)-\left(\sqrt{a-1}\right)\left(\sqrt{a-2}\right)+4-4\sqrt{a}}{\left(\sqrt{a-2}\right)\left(\sqrt{a+2}\right)}\)
\(Q=\frac{a+5\sqrt{a}+6-a+3\sqrt{a-2}+4-4\sqrt{a}}{\left(\sqrt{a-2}\right)\left(\sqrt{a+2}\right)}\)
\(Q=\frac{8+4\sqrt{a}}{\left(\sqrt{a-2}\right)\left(\sqrt{a+2}\right)}\)
\(Q=\frac{4\left(\sqrt{a+2}\right)}{\left(\sqrt{a+2}\right)\left(\sqrt{a-2}\right)}\)
\(Q=\frac{4}{\sqrt{a-2}}\)
26 tháng 4 2017
\(=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}\right)-2}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\frac{4}{a}}\)
\(=\frac{2a}{\sqrt{a-4}}\)
14 tháng 10 2020
a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)
\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)
b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)
\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)
c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
\(P=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)
\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\frac{4\sqrt{a}-4}{a-4}\)
\(=\frac{a+5\sqrt{a}+6-\left(a-3\sqrt{a}+2\right)-\left(4\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)