\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
\(\left(\sqrt{45}-2\sqrt{10}+\sqrt{5}\right)\sqrt{5}+5\sqrt{8}\)
Tính
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\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}=14\sqrt{3}\)
câu tiếp tương tự câu thứ 2 nha
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
\(a,3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\\ =6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\\ =14\sqrt{3}\)
\(b,\left(\sqrt{45}-2\sqrt{10}+\sqrt{5}\right)\sqrt{5}+5\sqrt{8}\\ =\sqrt{225}-10\sqrt{2}+5+10\sqrt{2}\\ =15+5=20\)
a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}\)
b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)
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
a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
= \(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)
= \(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)
= -2
b); c); d) làm tương tự
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: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
Bài làm:
1) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
\(=3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
\(=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
\(=14\sqrt{3}\)
2) \(\left(\sqrt{45}-2\sqrt{10}+\sqrt{5}\right).\sqrt{5}+5\sqrt{8}\)
\(=3\sqrt{5}.\sqrt{5}-2\sqrt{10}.\sqrt{5}+\sqrt{5}.\sqrt{5}+5.2\sqrt{2}\)
\(=15-10\sqrt{2}+5+10\sqrt{2}\)
\(=20\)