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\(a,\dfrac{7}{\sqrt{12}}=\dfrac{7\sqrt{3}}{\sqrt{12}\cdot\sqrt{3}}\)
\(=\dfrac{7\sqrt{3}}{\sqrt{36}}=\dfrac{7\sqrt{3}}{6}\)
\(b,\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}}{2\cdot3}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
\(c,\dfrac{1}{5\sqrt{12}}=\dfrac{\sqrt{3}}{5\cdot2\sqrt{3}\cdot\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{10\cdot3}=\dfrac{\sqrt{3}}{30}\)
\(d,\dfrac{2\sqrt{3}+3}{4\sqrt{3}}=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{4\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}}{4}\)
a) \(\dfrac{7}{\sqrt[]{12}}=\dfrac{7}{2\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{2\sqrt[]{3}.\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{6}\)
b) \(\dfrac{3}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}.\sqrt[]{3}}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{2}\)
c) \(\dfrac{1}{5\sqrt[]{12}}=\dfrac{1}{10\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{10\sqrt[]{3}.\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{30}\)
d) \(\dfrac{2\sqrt[]{3}+3}{4\sqrt[]{3}}=\dfrac{\sqrt[]{3}\left(2\sqrt[]{3}+3\right)}{4\sqrt[]{3}.\sqrt[]{3}}=\dfrac{3\left(2+\sqrt[]{3}\right)}{12}=\dfrac{2+\sqrt[]{3}}{4}\)
\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)
\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)
\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)
\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)
\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)
mấy bài dạng này bn nên sử dụng cách nhân liên hợp hoặc phân tích đa thức thành nhân tử nha . mk lm 1 bài còn lại thì bn tự lm cho quen nha :)
a) ta có : \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}-\sqrt{7}}=\dfrac{\left(\sqrt{6}+\sqrt{14}\right)\left(2\sqrt{3}+\sqrt{7}\right)}{\left(2\sqrt{3}-\sqrt{7}\right)\left(2\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{6\sqrt{2}+\sqrt{42}+2\sqrt{42}+7\sqrt{2}}{\left(2\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}=\dfrac{13\sqrt{2}+3\sqrt{42}}{5}\)
gợi ý : b) phân tích đa thức thành nhân tử bằng cách sử dụng hằng đẳng thức số \(6\)
c) nhân liên hợp 2 lần nha .
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}-\sqrt{7}}\)
=\(\dfrac{\left(\sqrt{6}+\sqrt{14}\right)\left(2\sqrt{3}+\sqrt{7}\right)}{\left(2\sqrt{3}-\sqrt{7}\right).\left(2\sqrt{3}+\sqrt{7}\right)}\)
=\(\dfrac{\left(\sqrt{6}+\sqrt{14}\right).\left(2\sqrt{3}+\sqrt{7}\right)}{12-7}\)
=\(\dfrac{2\sqrt{18}+\sqrt{42}+2\sqrt{42}+\sqrt{98}}{5}\)
=\(\dfrac{6\sqrt{2}+\sqrt{42}+2\sqrt{42}+7\sqrt{2}}{5}\)
=\(\dfrac{3\sqrt{42}+13\sqrt{2}}{5}\)
b) \(\dfrac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
=\(\dfrac{\left(5\sqrt{5}+3\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)}\)
=\(\dfrac{25-5\sqrt{15}+3\sqrt{15}-9}{2}\)
=\(\dfrac{16-2\sqrt{15}}{2}=8-\sqrt{15}\)
Câu c mk chưa làm được
\(\dfrac{3}{5-\sqrt{3}}=\dfrac{3\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}\)
\(=\dfrac{3\left(5+\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{3\left(5+\sqrt{3}\right)}{25-3}=\dfrac{3\left(5+\sqrt{3}\right)}{22}\)