Đặt \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=a\)
Thì ta có a3 = 4 - 3a
<=>( a3 - 1) + (3a - 3) = 0
<=> (a - 1)(a2 + a +4) = 0
<=> a = 1
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Đặt \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=a\)
Thì ta có a3 = 4 - 3a
<=>( a3 - 1) + (3a - 3) = 0
<=> (a - 1)(a2 + a +4) = 0
<=> a = 1
\(cm:\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(\)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) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
thực hiện phép tính: a)\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}+\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
b)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
c)\(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
d)\(\left(2\sqrt{5}+\sqrt{12}\right)\left(\sqrt{3}-\sqrt{5}\right)\)
e)\(\sqrt{2}+\sqrt{\frac{1}{2}}+\sqrt{72}-\sqrt{\frac{3}{2}}\)
f)\(\sqrt{2}\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
g)\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\left(2\sqrt{3}-2007\right)\)
Toán số 9, liên quan đến dạng căn. Các bạn giúp mình nhé, xin cảm ơn rất nhiều. :)
1) \(\left(\sqrt{3-2\sqrt{\sqrt{3}-1}}+\frac{\sqrt{3}-1}{\sqrt{2}}\right)\sqrt{\sqrt{3}-1}\)
2) \(\left(\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}\right):2\sqrt{\sqrt{5}-2}\)
3) \(\frac{\sqrt{10+6\sqrt{2}}-\sqrt{10-6\sqrt{2}}}{\sqrt{5-\sqrt{7}}}-\sqrt{9+2\sqrt{20}}\)
4) \(\frac{\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}}{\sqrt{5+\sqrt{22}}}-\frac{\sqrt{6-\sqrt{24}}}{\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}}\)
5) \(\sqrt{5+2\sqrt{14\sqrt{5}-26}}-\sqrt{4\sqrt{5}-1+\sqrt{80-8\sqrt{5}}}\)
6) \(\frac{\sqrt{11+\sqrt{5}}+\sqrt{11-\sqrt{5}}}{\sqrt{11+2\sqrt{29}}}-\sqrt{3-2\sqrt{2}}\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
Tính:
1) \(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}\)
2) \(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
3) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
4) \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{\sqrt{5}-3}\)
5) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}-\dfrac{1}{\sqrt{6}+\sqrt{2}}\)
LM CHI TIẾT GIÚP MK NHÉ
11) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) + \(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
12) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}\) + \(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\) - \(\dfrac{1}{2-\sqrt{3}}\)
Tính:
1) \(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
2) \(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}\)
3) \(\dfrac{\sqrt{2}+\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)
4) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
5) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
Rút gọn các biểu thức sau:
a \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
b \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
d \(\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)