\(\sqrt{28}+\sqrt{125}-3\sqrt{343}-\frac{3}{8}\sqrt{396}\)
b/\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
cần gấp
Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.
** Bài toán rút gọn**
Lời giải:
\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)
\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)
\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)
\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)
\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)
--------------------
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)
\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)
\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)
----------------------
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)
\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)
\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|-\left|4-2\sqrt{2}\right|=3-2\sqrt{2}-4+2\sqrt{2}\)
\(=-1\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|=3-2\sqrt{2}+3+2\sqrt{2}\)
\(=6\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
c) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
Rút Gọn
1.\(3\sqrt{2}-4\sqrt{18}-2\sqrt{32}-\sqrt{50}\)
2,\(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
3,\(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
4,\(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
5,\(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}\)
6,\(10\sqrt{28}-2\sqrt{275}-3\sqrt{343}-\frac{3}{2}\sqrt{396}\)
Rút Gọn
1.\(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
2.\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
3.\(\sqrt{72}+\sqrt{4\frac{1}{2}}-\sqrt{32}-\sqrt{162}\)
4.\(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)
5.\(\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\)
6.\(\left(2\sqrt{112}-5\sqrt{7}+2\sqrt{63}-2\sqrt{28}\right)\sqrt{7}\)
7.\(\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)\)
8.\(7\sqrt{24}-\sqrt{150}-5\sqrt{54}\)
9.\(2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
10.\(\sqrt{32}-\sqrt{50}+\sqrt{98}-\sqrt{72}\)
11.\(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
12.\(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
13.\(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
14.\(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
15.\(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}\)
16.\(10\sqrt{28}-2\sqrt{275}-3\sqrt{343}-\frac{3}{2}\sqrt{396}\)
Tính
a)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
b)\(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
c) \(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
d)\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
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 .
Rút gọn biểu thức.
a) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
b) \(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}\)
a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)
\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)
\(=2\left(3-2\sqrt{2}\right)=6-4\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}\)
`\sqrt(17-3\sqrt32)+\sqrt(17+3\sqrt32)`
`=\sqrt(17-12\sqrt2)+\sqrt(17+12\sqrt2)`
`=\sqrt(9-12\sqrt2+8)+\sqrt(9+12\sqrt2+8)`
`=\sqrt(3^2-2.3.2\sqrt2 +(2\sqrt2)^2)+\sqrt(3^2+2.3.2\sqrt2+(2\sqrt2)^2)`
`=\sqrt((3-2\sqrt2)^2)+\sqrt((3+2\sqrt2)^2)`
`=3-2\sqrt2+3+2\sqrt2`
`=6`
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
a) \(\sqrt{2}+\frac{1}{\sqrt{5+2\sqrt{6}}}+\frac{2}{\sqrt{8+2\sqrt{15}}}\)
b) \(\frac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}+\frac{3+2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\frac{1}{\sqrt{12+2\sqrt{35}}}\)
c) \(\left(\frac{15}{3-\sqrt{2}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
Giúp mình bài này nhé, mình đang cần gấp mọi người ơi :<
Tính
a,\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
b,\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
c,\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
d,\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}\)
\(=2\sqrt{5}+4\sqrt{2}\)