\(x^3=\left(\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\right)^3=\sqrt[3]{5+2\sqrt{6}}^3\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)^2}.\sqrt[3]{5-2\sqrt{6}}+3\sqrt[3]{5+2\sqrt{6}}.\sqrt[3]{\left(5-2\sqrt{6}\right)^2}+\sqrt[3]{5-2\sqrt{6}}^3\)
\(=5+2\sqrt{6}+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5+2\sqrt{6}}\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5-2\sqrt{6}}+5-2\sqrt{6}\)
\(=5+5+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5-2\sqrt{6}}\)
\(=10+ 3\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{5-2\sqrt{6}}\)
p/s : có bạn hỏi nên mình đăng , các bạn đừng report nhé
11) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) + \(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
12) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}\) + \(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\) - \(\dfrac{1}{2-\sqrt{3}}\)
\(\)1) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\)
2) \(\dfrac{2\sqrt{6}-\sqrt{10}}{4\sqrt{3}-2\sqrt{5}}\)
3) \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
4) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
2+√5/√2+√(3+√5)+2-√5//√2-√(3+√5)
rút gọn
g, \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right).\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\) h,\(\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{3\dfrac{1}{3}}\right).\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\dfrac{1}{5}}\right)\)
C= 2+√5/√2+√(3+√5)+2-√5/√2-√(3-√5)
Q=(5-2 căn 5/2-căn 5 trừ 2 riêng).(5+3 căn 5/3+ căn 5 trừ 2 riêng)
\(a:\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
b : \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
c : \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right).\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
d : \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
Toán số 9, liên quan đến dạng căn. Các bạn giúp mình nhé, xin cảm ơn rất nhiều. :)
1) (√3−2√√3−1+√3−1√2 )√√3−1
2) (√5+2√9√5−19−√7−√5):2√√5−2
3) √10+6√2−√10−6√2√5−√7 −√9+2√20
4) √5+√3+√5−√3√5+√22 −√6−√24√3+√3−√3−√3
5) √5+2√14√5−26−√4√5−1+√80−8√5
6) √11+√5+√11−√5√11+2√29 −√3−2√2
