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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) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)
\(=\sqrt{\left(4-3\right)\cdot4}\)
\(=\sqrt{1\cdot4}\)
\(=\sqrt{4}\)
\(=2\)
b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)
\(=0+14\)
\(=14\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
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: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
a) bấm mày
b) qui đồng trong ngặc trước rồi thu gọn
c) trong ngặc : khử phân số thứ nhất \(\Rightarrow\) qui đồng \(\Rightarrow\) giải bình thường
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
a) \(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
= \(\left(5+4\sqrt{2}\right)\left(9-\left(2\sqrt{1+\sqrt{2}}\right)^2\right)\)
= \(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\) = \(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
= \(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)\) = \(25-\left(4\sqrt{2}\right)^2\) = \(25-32=-7\)