giai hộ với :
\(\frac{12}{3-\sqrt{3}}\)
K
Khách
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PN
29 tháng 7 2020
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
TN
0
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2019
ĐKXĐ: \(x\ge0;x\ne1\)
\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+19\sqrt{x}-14}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
TM
0
HA
13 tháng 10 2015
bài 3 :
nhân đảo ngược căn 2 - căn 3 rồi quy đồng là ra ngay
HT
2 tháng 8 2015
a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)
b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)
c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\Leftrightarrow\sqrt{3}x=4\sqrt{3}\Leftrightarrow x=4\)
20 tháng 9 2020
Ta có: \(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}}{3}+\frac{\sqrt{2}}{3}+\frac{2\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12}.\sqrt{\frac{1}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{5-2\sqrt{6}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left(\sqrt{3}-\sqrt{2}\right)\)(vì \(\sqrt{3}-\sqrt{2}>0\))
\(=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{3}=\sqrt{3}\)
DH
15 tháng 11 2019
\(\frac{4}{\sqrt{7}-\sqrt{3}}+\frac{6}{3+\sqrt{3}}+\frac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\frac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\sqrt{7}\left(1-\sqrt{7}\right)}{\sqrt{7}-1}\)
\(=\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{6\left(3-\sqrt{3}\right)}{9-3}-\sqrt{7}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
\(=3\)
\(\frac{2.6}{3-\sqrt{3}}\) = \(\frac{2\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{3-\sqrt{3}}\)= \(2\left(3+\sqrt{3}\right)\)