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\(1,\sqrt{432}-\sqrt{363}+\sqrt{48}-\sqrt{75}+\sqrt{108}-\sqrt{147}\)
\(=\sqrt{12^2.3}-\sqrt{11^2.3}+\sqrt{4^2.3}-\sqrt{5^2.3}+\sqrt{6^2.3}-\sqrt{7^2.3}\)
\(=12\sqrt{3}-11\sqrt{3}+4\sqrt{3}-5\sqrt{3}+6\sqrt{3}-7\sqrt{3}\)
\(=\sqrt{3}.\left(12-11+4-5+6-7\right)\)
\(=-\sqrt{3}\)
\(2,6\sqrt{60}-5\sqrt{8}+3\sqrt{15}+4\sqrt{32}+3\sqrt{128}-2\sqrt{1250}\)
\(=6.2\sqrt{15}-5.2\sqrt{2}+3\sqrt{15}+4.4\sqrt{2}+3.8\sqrt{2}-2.25\sqrt{2}\)
\(=12\sqrt{15}+3\sqrt{15}-10\sqrt{2}+16\sqrt{2}+24\sqrt{2}-50\sqrt{2}\)
\(=\sqrt{15}.\left(12+3\right)+\sqrt{2}.\left(-10+16+24-50\right)\)
\(=15\sqrt{15}-20\sqrt{2}\)
1/ \(\sqrt{432}-\sqrt{363}+\sqrt{48}-\sqrt{75}+\sqrt{108}-\sqrt{147}\)
\(=12\sqrt{3}-11\sqrt{3}+4\sqrt{3}-5\sqrt{3}+6\sqrt{3}-7\sqrt{3}\)
\(=\left(12-11+4-5+6-7\right)\sqrt{3}\)
\(=-\sqrt{3}\)
2/ \(6\sqrt{60}-5\sqrt{8}+3\sqrt{15}+4\sqrt{32}+3\sqrt{128}-2\sqrt{1250}\)
\(=12\sqrt{15}-10\sqrt{2}+3\sqrt{15}+16\sqrt{2}+24\sqrt{2}-50\sqrt{2}\)
\(=\left(12+3\right)\sqrt{15}+\left(-10+16+24-50\right)\sqrt{2}\)
\(=15\sqrt{15}-20\sqrt{2}\)
7+8=98. Vì 7+8=7.6+8.7=98
1+1+1+1+1+1+1+1+1+1+1+1.0+1=0+1=1
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
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: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)
Bài làm của: Phùng Khánh Linh
c)\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
= \(\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{4^2-2.4.\sqrt{8}+\left(\sqrt{8}\right)^2}\)
= \(\sqrt{\left(3-2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{\left(4-\sqrt{8}\right)^2}\)
= \(\left|3-2\sqrt{2}\right|-\left|4-\sqrt{8}\right|\)
= (3 - 2\(\sqrt{2}\)) - (4 - \(\sqrt{8}\))
= 3 - 2\(\sqrt{2}\) - 4 + \(\sqrt{8}\)
= -1
\(a.\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\text{|}\sqrt{3}+1\text{|}-\text{|}\sqrt{3}-1\text{|}=2\)\(b.\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\text{|}\sqrt{5}-2\text{|}-\text{|}\sqrt{5}+2\text{|}=-4\) Còn lại tương tự nhé .
1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)
7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)
8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)
9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)
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 !
\(3\sqrt{8}-4\sqrt{2}+5\sqrt{32}\\ =3\sqrt{4.2}-4\sqrt{2}+5\sqrt{16.2}\\ =3.2\sqrt{2}-4\sqrt{2}+5.4\sqrt{2}\\ =6\sqrt{2}-4\sqrt{2}+20\sqrt{2}\\ =22\sqrt{2}\)