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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= (\(\frac{3\sqrt{a}}{\sqrt{a}+4}+\frac{\sqrt{a}}{\sqrt{a}-4}+\frac{4\left(a+2\right)}{16-a}\)):\(\left(1-\frac{2\sqrt{a}+5}{\sqrt{a}-4}\right)\)
=\(\left(\frac{3\sqrt{a}\left(\sqrt{a}-4\right)}{a-16}+\frac{\sqrt{a}\left(\sqrt{a}+4\right)}{a-16}-\frac{4a+8}{a-16}\right):\left(\frac{\sqrt{a}-4-2\sqrt{a}-5}{\sqrt{a}-4}\right)\)
= \(\left(\frac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{a-16}\right):\left(\frac{-\sqrt{a}-9}{\sqrt{a}-4}\right)\)
=\(\left(\frac{-8\sqrt{a}-8}{a-16}\right).\left(\frac{\sqrt{a}-4}{-\sqrt{a}-9}\right)=\frac{8\sqrt{a}+8}{\left(\sqrt{a}+4\right).\left(\sqrt{a}+9\right)}=\frac{8\sqrt{a}+8}{a+13\sqrt{a}+36}\)
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) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
18 tháng 7 2016
\(P=\frac{1}{1-\sqrt{a}}+\frac{a\sqrt{a}}{\sqrt{a}-1}\)
\(=\frac{-1}{\sqrt{a}-1}+\frac{a\sqrt{a}}{\sqrt{a}-1}\)
\(=\frac{a\sqrt{a}-1}{\sqrt{a}-1}\)
18 tháng 7 2016
Với a = 1/4 thì:
\(P=\frac{a\sqrt{a}-1}{\sqrt{a}-1}=\frac{\frac{1}{4}\sqrt{\frac{1}{4}}-1}{\sqrt{\frac{1}{4}}-1}=\frac{\frac{1}{4}\cdot\frac{1}{2}-1}{\frac{1}{2}-1}=\frac{\frac{1}{8}-1}{\frac{1}{2}-1}=\frac{1-\frac{1}{8}}{1-\frac{1}{2}}=\frac{\frac{7}{8}}{\frac{1}{2}}=\frac{7}{4}\)
HH
17 tháng 8 2016
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
17 tháng 11 2017
a, = \(\sqrt{a^2b^2.\left(1+\frac{1}{a^2b^2}\right)}\) = \(\sqrt{a^2b^2+1}\)
c, = \(\sqrt{\frac{a+ab}{b^4}}\) = \(\frac{\sqrt{a+ab}}{b^2}\)
k mk nha
17 tháng 11 2017
a, \(ab\sqrt{1+\frac{1}{a^2b^2}}\)
\(ab\sqrt{1+\frac{1}{a^2b^2}}=ab\sqrt{\frac{1+a^2b^2}{a^2b^2}}=\frac{ab}{\left|ab\right|}\sqrt{1+a^2b^2}\)
\(=\hept{\begin{cases}\sqrt{1+a^2b^2}ĐK:ab>0\\-\sqrt{1+a^2b^2}ĐKab< 0\end{cases}}\)
b, \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}\)
\(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a+ab}{b^4}}=\frac{1}{b^2}\sqrt{a+ab}\)
\(\frac{1}{a-4}-\frac{1}{a+4\sqrt{a}+4}\)
\(\frac{1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{1}{\left(\sqrt{a}+2\right)^2}\)
\(\frac{\sqrt{a}+2-\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}\)
\(\frac{\sqrt{a}+2-\sqrt{a}+2}{a\sqrt{a}+4a+4\sqrt{a}-2a-8\sqrt{a}-8}\)
\(\frac{4}{a\sqrt{a}+2a-4\sqrt{a}-8}\)