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BT
2 tháng 8 2017
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
11 tháng 8 2019
b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))
t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).\(\sqrt{2}\)(\(\sqrt{5}\) -1)
t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))
t = (\(\sqrt{5}\) -1)2.(3 +\(\sqrt{5}\))
t = (5 - \(2\sqrt{5}\)+1).(3 +\(\sqrt{5}\))
t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)
t = 6
MA
3 tháng 1 2019
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
TB
2
NC
29 tháng 8 2020
a) \(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(A=\sqrt{16+8\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}\)
\(A=\sqrt{\left(4+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(A=4+\sqrt{3}-\sqrt{3}-1=3\)
b) \(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(B=\sqrt{25+10\sqrt{2}+2}-\sqrt{16+8\sqrt{2}+2}\)
\(A=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(A=5+\sqrt{2}-4-\sqrt{2}=1\)
29 tháng 8 2020
\(A=\sqrt{19+8\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3+8\sqrt{3}+16}-\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot4+4^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}+1^2}\)
\(=\sqrt{\left(\sqrt{3}+4\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}+4\right|-\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}+4-\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}+4-\sqrt{3}-1=3\)
\(B=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=\sqrt{2+10\sqrt{2}+25}-\sqrt{2+8\sqrt{2}+16}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot5+5^2}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot4+4^2}\)
\(=\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{\left(\sqrt{2}+4\right)^2}\)
\(=\left|\sqrt{2}+5\right|-\left|\sqrt{2}+4\right|\)
\(=\sqrt{2}+5-\left(\sqrt{2}+4\right)\)
\(=\sqrt{2}+5-\sqrt{2}-4=1\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 3 2019
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/ \(\sqrt{19+8\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(4+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=4+\sqrt{3}+2-\sqrt{3}=6\)
CW
1 tháng 8 2017
tớ ko chép lại đề, kí hiệu nhé
(1) \(=\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{\left|\sqrt{6}+\sqrt{5}\right|^2}=\left(\sqrt{6}-\sqrt{5}\right)^2-\left(\sqrt{6}+\sqrt{5}\right)=1-2\sqrt{30}-\sqrt{6}-\sqrt{5}\)
ai ra đề mà để đáp án dài thế này mất thẩm mĩ quá!!!
(2) \(=\sqrt{\left|\sqrt{5}+\sqrt{3}\right|^2}-\sqrt{\left|\sqrt{5}-\sqrt{3}\right|^2}=\left(\sqrt{5}+\sqrt{3}\right)-\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}\)
(3) \(=\sqrt{\left|\sqrt{7}+2\right|^2}-\sqrt{\left|3-\sqrt{5}\right|^2}=\sqrt{7}+2-3+\sqrt{5}=\sqrt{7}+\sqrt{5}-1\)
lại thêm 1 phép tính không đẹp....
(4) \(=\sqrt{\left|3\sqrt{2}-2\right|^2}-\sqrt{\left|3\sqrt{2}+1\right|^2}=3\sqrt{2}-2-3\sqrt{2}-1=-3\)
(5) \(=\sqrt{\left|2\sqrt{3}-1\right|^2}+\sqrt{\left|2\sqrt{3}-3\right|^2}=2\sqrt{3}-1+2\sqrt{3}-3=4\sqrt{3}-4\)
kiểm tra lại kết quả nhé ^^! Cảm ơn!
Y
18 tháng 7 2019
I: Rút gọn
\(A=\sqrt{7-4\sqrt{3}}\\ =\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}\\ =2-\sqrt{3}\)
\(B=\sqrt{19-8\sqrt{3}}\\ =\sqrt{16-2\cdot4\cdot\sqrt{3}+3}\\ =\sqrt{\left(4-\sqrt{3}\right)^2}\\ =4-\sqrt{3}\)
\(C=\sqrt{21-4\sqrt{5}}\\ =\sqrt{20-2\cdot2\sqrt{5}+1}\\ =\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot1+1}\\ =\sqrt{\left(2\sqrt{5}-1\right)^2}\\ =2\sqrt{5}-1\)
Câu D mình làm chưa ra, sorry :<
\(\sqrt{4-2\sqrt{3}}\)+\(\sqrt{19-8\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.4+4^2}\)
=\(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}-4\right)^2}\)\(=\sqrt{3}-1+4-\sqrt{3}=3\)