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1.\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}=3\sqrt{2}\)
2.\(=5\sqrt{5}+4\sqrt{5}-9\sqrt{5}=0\)
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
1,\(4\sqrt{5}+2\sqrt{5}-\sqrt{5}-15\sqrt{5}=-10\sqrt{5}\)
2,\(8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
3,\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}=33\)
4,\(7\sqrt{7a}+3\sqrt{7a}-2\sqrt{7a}=8\sqrt{7a}\)
5,\(-6\sqrt{a}-\sqrt{6a}+\sqrt{6a}=-6\sqrt{a}\)
6,\(8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)
2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)
3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)
4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)
5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
bài 1
biểu thức có nghĩa khi x, y thỏa mãn đồng thời
\(\left\{{}\begin{matrix}x,y\ne0\\\dfrac{y}{x}\ge0\end{matrix}\right.\Rightarrow x.y>0}\)x, y khác 0
x.y>0
a: \(=\left(2\sqrt{3}-12\sqrt{3}+15\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
b: \(=\left(6\sqrt{2}-16\sqrt{2}+15\sqrt{2}\right):5=\sqrt{2}\)
c: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)