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Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)
b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)
c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)
d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)
f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
3: \(\sqrt{12-3\sqrt{7}}-\sqrt{12-3\sqrt{7}}=0\)
4: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=-2\sqrt{2}\)
6: \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-4\sqrt{3}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
1: \(=\sqrt{5}-2\)
2: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
4: \(=\sqrt{2}+1-2+\sqrt{2}=-1+2\sqrt{2}\)
5: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-\sqrt{5}-1}{2}=\dfrac{15-\sqrt{5}}{2}\)