TT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TX
22 tháng 8 2018
Câu 1:
a, \(\sqrt{50.98} = 5\sqrt{2} . 7\sqrt{2} = 70\)
b, \(\sqrt{2,5.12,1} = 30,25\)
c, \(\sqrt{17.51.27} = \sqrt{23409} = 153\)
d, \(\sqrt{32.128} = \sqrt{4096} = 64\)
e, \(\sqrt{3,2.7,2.49} = 7\sqrt{3,2.7,2} = 7\sqrt{23,04} =33,6\)
g, \(\sqrt{2,5.12,5.20} = \sqrt{625} = 25\)
12 tháng 7 2021
Bài 1:
a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)
b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)
Bài 2:
a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)
b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)
c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)
\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)
\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)
\(=\dfrac{6}{35}\)
21 tháng 5 2023
Tính:
a, √49 . √144+ √256 : √64
= 7 . 12 + 16 : 8
= 84 + 2
= 86
b, 72 : √2^2.36.3^2- √225
= 72: 2.6.3-15
= -13
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 7 2021
a.
\(=\sqrt{\sqrt{5}-2}-\sqrt{5\left(\sqrt{5}+2\right)}+2\sqrt{\sqrt{5}+2}\)
\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}+2}\left(\sqrt{5}-2\right)\)
\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}-2}\left(\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\right)\)
\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}-2}.1=0\)
b.
\(=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\left(\sqrt{2}+1\right)}\)
\(=\sqrt{\sqrt{2}-1}-\left(\sqrt{2}-1\right)\left(\sqrt{\sqrt{2}+1}\right)\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}.\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}=0\)
H
9 tháng 10 2023
\(A=\sqrt{243}-\sqrt{27}+\sqrt{3}-\sqrt{48}\\ =\sqrt{81\cdot3}-\sqrt{9\cdot3}+\sqrt{3}-\sqrt{16\cdot3}\\ =9\sqrt{3}-3\sqrt{3}+\sqrt{3}-4\sqrt{4}\\ =\left(9-3+1-4\right)\sqrt{3}\\ =3\sqrt{3}\)
\(B=\dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{3}+\sqrt{5}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\sqrt{5}-\sqrt{3}\\ =\sqrt{5}+1+\sqrt{3}-\sqrt{5}-\sqrt{3}\\ =1\)
NH
21 tháng 5 2023
a)
\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )
\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)
b)
\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )
\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)
26 tháng 9 2021
a)\(\sqrt{\dfrac{2}{2-\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)\(=\sqrt{2\left(2+\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\sqrt{2^2.6}+\dfrac{2\sqrt{24}}{8}+\dfrac{1}{6}\)
\(=\dfrac{\sqrt{6}}{3}-2\sqrt{6}+\dfrac{\sqrt{2^2.6}}{4}+\dfrac{1}{6}=\dfrac{-5\sqrt{6}}{3}+\dfrac{2\sqrt{6}}{4}+\dfrac{1}{6}\)
\(=\dfrac{-7\sqrt{6}}{6}+\dfrac{1}{6}\)
26 tháng 9 2021
Nhầm xíu
\(\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\left|\sqrt{3}-1\right|=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
30 tháng 6 2021
a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)
\(=-3\cdot4\cdot3\sqrt{10}\)
\(=-36\sqrt{10}\)
b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)
\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)
\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)
\(=15\sqrt{3}\)
c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)
\(=4\cdot3-4-2\cdot2\)
\(=12-4-4=4\)