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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
c) \(\dfrac{3\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\dfrac{3\sqrt{3}}{\left(\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}\right)}=\dfrac{3\sqrt{3}\left(\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}\right)}{\left(\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}\right)\left(\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}\right)}\) = \(\dfrac{3\sqrt{6}+9-3\sqrt{15}}{\left(\sqrt{2}+\sqrt{3}\right)^2-5}\) = \(\dfrac{3\sqrt{6}+9-3\sqrt{15}}{2+2\sqrt{6}+3-5}=\dfrac{3\sqrt{6}+9-3\sqrt{15}}{2\sqrt{6}}\)
= \(\dfrac{\left(3\sqrt{6}+9-3\sqrt{15}\right)\sqrt{6}}{2\sqrt{6}.\sqrt{6}}\) = \(\dfrac{18+9\sqrt{6}-9\sqrt{10}}{12}\)
= \(\dfrac{3\left(6+3\sqrt{6}-3\sqrt{10}\right)}{3.4}=\dfrac{6+3\sqrt{6}-3\sqrt{10}}{4}\)
d) \(\dfrac{4}{1+\sqrt{2}+\sqrt{3}}=\dfrac{4}{\left(\left(1+\sqrt{2}\right)+\sqrt{3}\right)}=\dfrac{4\left(\left(1+\sqrt{2}\right)-\sqrt{3}\right)}{\left(\left(1+\sqrt{2}\right)+\sqrt{3}\right)\left(\left(1+\sqrt{2}\right)-\sqrt{3}\right)}\)
= \(\dfrac{4+4\sqrt{2}-4\sqrt{3}}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{4+4\sqrt{2}-4\sqrt{3}}{1+2\sqrt{2}+1-3}\) = \(\dfrac{4+4\sqrt{2}-4\sqrt{3}}{2\sqrt{2}}\)
\(\dfrac{\left(4+4\sqrt{2}-4\sqrt{3}\right)\sqrt{2}}{2\sqrt{2}\sqrt{2}}=\dfrac{4\sqrt{2}+8-4\sqrt{6}}{4}\) = \(\dfrac{4\left(\sqrt{2}+4-\sqrt{6}\right)}{4}=\sqrt{2}+4-\sqrt{6}\)
câu a thôi nha
câu b:\(\dfrac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}{12}=\dfrac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\)
câu c,d tương tự câu b thôi
bản chất lười =))
câu e mình viết sai đề, mk sửa lại nhé , với mình bổ sung câu f
e) \(\dfrac{2}{\sqrt[3]{4}+\sqrt[3]{5}}\)
f) \(\dfrac{1}{2-\dfrac{\sqrt[3]{3}}{2}}\)
Bài 50:
\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)
\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{5-2\sqrt{5}+1}{5-1}=\dfrac{2\left(3-\sqrt{5}\right)}{4}=\dfrac{3-\sqrt{5}}{2}\)
b: \(\dfrac{37}{7+2\sqrt{3}}=7-2\sqrt{3}\)
c:\(=\dfrac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{5}\right)}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\)
d: \(=\dfrac{\left(1+\sqrt{a}\right)\cdot\left(2+\sqrt{a}\right)}{4-a}\)
bài 2:
a: \(\dfrac{25}{5-2\sqrt{3}}=\dfrac{125+10\sqrt{3}}{13}\)
b: \(\dfrac{8}{\sqrt{5}+2}=8\sqrt{5}-32\)
c: \(\dfrac{6}{2\sqrt{3}-\sqrt{7}}=\dfrac{12\sqrt{3}+6\sqrt{7}}{5}\)
d: \(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\dfrac{\sqrt{6}}{2}\)
a: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{7+2\sqrt{10}-3}=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{4+2\sqrt{10}}\)
\(=\dfrac{-\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(4-2\sqrt{10}\right)}{24}\)
b: \(=\dfrac{2+\sqrt{3}+\sqrt{5}}{4-8+2\sqrt{15}}=\dfrac{2+\sqrt{3}+\sqrt{5}}{2\sqrt{15}-4}\)
\(=\dfrac{\left(2+\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+4\right)}{44}\)
a. \(\dfrac{1}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}\right)^2-2}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{5+3-2-2\sqrt{15}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{6-2\sqrt{15}}=\dfrac{\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{15}\right)}{\left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right)2}=\dfrac{3\sqrt{5}-3\sqrt{3}-3\sqrt{2}+5\sqrt{3}-3\sqrt{5}-\sqrt{30}}{\left(9-15\right).2}=\dfrac{2\sqrt{3}-3\sqrt{2}-\sqrt{30}}{-12}\)b. \(\dfrac{1}{2-\sqrt{3}-\sqrt{5}}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}-\sqrt{5}\right)\left(2-\sqrt{3}+\sqrt{5}\right)}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}\right)^2-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{4-4\sqrt{3}+3-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{2-4\sqrt{3}}=\dfrac{\left(2-\sqrt{3}+\sqrt{5}\right)\left(1+2\sqrt{3}\right)}{2\left(1-2\sqrt{3}\right)\left(1+2\sqrt{3}\right)}=\dfrac{2+4\sqrt{3}-\sqrt{3}-6+\sqrt{5}+2\sqrt{15}}{2.\left(1-12\right)}=\dfrac{3\sqrt{3}+\sqrt{5}+2\sqrt{15}-4}{-22}\)
a) Ta có: \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{1-2\sqrt{15}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{-7\left(\sqrt{5}+10\sqrt{3}-\sqrt{3}-6\sqrt{5}-\sqrt{7}-2\sqrt{105}\right)}{59}\)
\(=\dfrac{-7\left(-5\sqrt{5}+9\sqrt{3}-\sqrt{7}-2\sqrt{105}\right)}{59}\)