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a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
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
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: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
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
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: \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-4\sqrt{3}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
a: \(=\dfrac{1}{2}\cdot2\sqrt{3}+3\sqrt{3}-5\sqrt{3}=-\sqrt{3}\)
b: \(=2-\sqrt{3}-\sqrt{3}-1=1\)
c: \(=18\sqrt{3}-10\sqrt{3}-\dfrac{1}{2}\cdot10\sqrt{3}=3\sqrt{3}\)
d: \(=\sqrt{10}+\sqrt{3}-\sqrt{5}+\sqrt{2}-2\sqrt{3}=\sqrt{10}+\sqrt{2}-\sqrt{3}-\sqrt{5}\)