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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, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
MK CHỈ VIẾT KQ THUI NHA ! VÌ DÀI QUÁ ...
A= 4,236067977 B = 2,414213562 C= 0,8218544151
D= 3,968118785 E= \(-\)\(10\sqrt{2}\) F=17,10050878 (\(3\sqrt{5}+6\sqrt{3}\))
G=\(-7\sqrt{5}\) H= \(-10\sqrt{2}\)
K VÀ KB NHOA ! Dương Nguyễn Ngọc Khánh !
1,
\(2\sqrt{5}-\sqrt{125}-\sqrt{80}\\ =2\sqrt{5}-\sqrt{25\cdot5}-\sqrt{16\cdot5}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}\\ =-7\sqrt{5}\)
2,
\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\\ =3\sqrt{2}-\sqrt{4\cdot2}+\sqrt{25\cdot2}-4\sqrt{16\cdot2}\\ =3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}\\=-10\sqrt{2}\)
3,
\(\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =\sqrt{9\cdot2}-3\sqrt{16\cdot5}-2\sqrt{25\cdot2}+2\sqrt{9\cdot5}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)
4,
\(\sqrt{27}-2\sqrt{3}+2\sqrt{48}-3\sqrt{75}\\ =\sqrt{9\cdot3}-2\sqrt{3}+2\sqrt{16\cdot3}-3\sqrt{25\cdot2}\\ =3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}\\ =-6\sqrt{3}\)
5,
\(3\sqrt{2}-4\sqrt{18}+\sqrt{32}-\sqrt{50}\\ =3\sqrt{2}-4\sqrt{9\cdot2}+\sqrt{16\cdot2}-\sqrt{25\cdot2}\\ =3\sqrt{2}-12\sqrt{2}+4\sqrt{2}-5\sqrt{2}\\ =-10\sqrt{2}\)
6,
\(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\\ =2\sqrt{3}-\sqrt{25\cdot3}+2\sqrt{4\cdot3}-\sqrt{49\cdot3}\\ =2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\\ =-6\sqrt{3}\)
7,
\(\sqrt{20}-2\sqrt{45}-3\sqrt{80}+\sqrt{125}\\ =\sqrt{4\cdot5}-2\sqrt{9\cdot5}-3\sqrt{16\cdot5}+\sqrt{25\cdot5}\\ =2\sqrt{5}-6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\\ =-11\sqrt{5}\)
8,
\(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =6\sqrt{4\cdot3}-\sqrt{4\cdot5}-2\sqrt{9\cdot3}+\sqrt{25\cdot5}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\\ =3\left(2\sqrt{3}+\sqrt{5}\right)\)
9,
\(4\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\\ =4\sqrt{4\cdot6}-2\sqrt{9\cdot6}+3\sqrt{6}-\sqrt{25\cdot6}\\ =8\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}=0\)
10,
\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)
a)\(2\sqrt{18}+3\sqrt{8}-3\sqrt{32}-\sqrt{50}\)
\(=2\sqrt{9.2}+3\sqrt{4.2}-3\sqrt{16.2}-\sqrt{25.2}\)
\(=6\sqrt{2}+6\sqrt{2}-12\sqrt{2}-5\sqrt{2}\)
\(=-5\sqrt{2}\)
b) \(\sqrt{200}-\sqrt{32}-\sqrt{72}\)
\(=\sqrt{100.2}-\sqrt{16.2}-\sqrt{36.2}\)
\(=10\sqrt{2}-4\sqrt{2}-6\sqrt{2}\)
\(=0\)
c) \(\sqrt{175}-\sqrt{112}+\sqrt{63}\)
\(=\sqrt{25.7}-\sqrt{16.7}+\sqrt{9.7}\)
\(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}\)
\(=4\sqrt{7}\)
d) \(3\sqrt{8}-\sqrt{32}+4\sqrt{2}+\sqrt{162}\)
\(=3\sqrt{4.2}-\sqrt{16.2}+4\sqrt{2}+\sqrt{81.2}\)
\(=6\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}\)
\(=15\sqrt{2}\)
a/ \(2\sqrt{10}-10\sqrt{10}+9\sqrt{10}=\sqrt{10}\)
b/ \(\frac{-1\left(4-3\sqrt{2}\right)+1\left(4+3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\frac{-4+3\sqrt{2}+4+3\sqrt{2}}{16-18}=\frac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c/ \(\left(3+\sqrt{5}\right).\sqrt{2}.\sqrt{7-3\sqrt{5}}=\left(3+\sqrt{5}\right)\sqrt{14-6\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\sqrt{\left(3-\sqrt{5}\right)^2}=\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
d/ \(3\sqrt{2}-4\sqrt{2}+5\sqrt{2}=4\sqrt{2}\)
e/ \(\sqrt{19+8\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(4+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=4+\sqrt{3}+2-\sqrt{3}=6\)
\(a,\left(\sqrt{27}-2\sqrt{17}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=3\sqrt{21}-2\sqrt{119}+7+7\sqrt{8}\)
Đề sai chăng???
\(b,\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{2}-1+\sqrt{2}+1\)
\(=2\sqrt{2}\)
\(c,9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)
\(=9\sqrt{2}-8\sqrt{2}-5\sqrt{2}+8\sqrt{2}\)
\(=\sqrt{2}\left(9-8-5+8\right)\)
\(=4\sqrt{2}\)
\(d,\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4+2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1-2-\sqrt{2}\)
\(=-3\)
a) \(2\sqrt{50}-3\sqrt{32}-\sqrt{162}+5\sqrt{98}\)
=\(2.5\sqrt{2}-3.4\sqrt{2}-9\sqrt{2}+5.7\sqrt{2}\)
= \(10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}\)
= \(24\sqrt{2}\)
b) \(\sqrt{8+2\sqrt{7}}+\sqrt{11-4\sqrt{7}}\)
= \(\sqrt{7+2\sqrt{7}+1}+\sqrt{7-4\sqrt{7}+4}\)
= \(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-2\right)^2}\)
= \(\sqrt{7}+1+\sqrt{7}-2\)
= \(2\sqrt{7}-1\)
c) \(\dfrac{10}{\sqrt{5}}+\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{18}-3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
= \(2\sqrt{5}+6-2\sqrt{5}-3\)
= 3
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. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}.\)