a) \(\dfrac{5-2\sqrt{ }5}{\sqrt{ }5-2}-\dfrac{11}{4+\sqrt{ }5} \)
b)\(\sqrt{9+4\sqrt{ }5-\sqrt{ }6-2\sqrt{ }5}\)
c)\(\sqrt{17-3\sqrt{ }32+\sqrt{ }17+\sqrt{ }32}\)
B1:Tính
a,\(\sqrt{\left(4-\sqrt{17}\right)^2}-\left(\sqrt{17}+2\right)\) b,\(\dfrac{7}{\sqrt{3}-\sqrt{2}}-\sqrt{147}-2\sqrt{18}\)
c,\(\dfrac{6}{\sqrt{5}-2}-\dfrac{6}{\sqrt{5}+2}+\sqrt{8}-4\sqrt{\dfrac{1}{7}}\) ; \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)
\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{17-12\sqrt{2}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot\sqrt{7}+\left(\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{\left(2\sqrt{2}-3\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{\left(3-2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{2\sqrt{2}-3}+\dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}-\dfrac{1}{\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(2\sqrt{2}+3\right)^2}{\left(2\sqrt{2}+3\right)\left(2\sqrt{2}-3\right)}+1-\dfrac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=17-12\sqrt{2}+1-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{2\cdot\left(18-12\sqrt{2}\right)}{2}-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{36-24\sqrt{2}-\sqrt{5}+\sqrt{7}}{2}\)
1.Rút gọn các BT sau:
a) 2\(\sqrt{3}\) +\(\sqrt{\left(2-\sqrt{3}\right)^2}\)
b)\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
c) \(\sqrt{17-2\sqrt{32}}\) +\(\sqrt{17+2\sqrt{32}}\)
a, \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)
b,\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{25-5}=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}=\dfrac{60}{20}=3\)
Câu 1: Rút gọn:
a) \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}\)
b) \(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
c) \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}\)
Câu 2: Giải phương trình:
\(\sqrt{9x^2-30x+25}=5\)
1.
a, \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}=6\sqrt{2}-20\sqrt{2}-12\sqrt{2}=-2\sqrt{2}\)
b, \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}+3\right)^2}\)
\(=\left|\sqrt{5}-3\right|+\left|\sqrt{5}+3\right|\)
\(=-\sqrt{5}+3+\sqrt{5}+3=6\)
c, \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
2.
ĐK: \(x\in R\)
\(\sqrt{9x^2-30x+25}=5\)
\(\Leftrightarrow\sqrt{\left(3x-5\right)^2}=5\)
\(\Leftrightarrow\left|3x-5\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=5\\3x-5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=0\end{matrix}\right.\)
Vậy ...
1.\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
2.\(\sqrt{9-4\sqrt{5}}-\sqrt{9+\sqrt{80}}\)
3.\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
4.\(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
5.\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
6.\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
7.\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
2.\(\sqrt{9-4\sqrt{5}}-\sqrt{9+\sqrt{80}}\)
\(=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
Tính :
a. \(\sqrt{\left(\sqrt{3}-3\right)^2}-\sqrt{16+6\sqrt{3}}\)
b. \(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{5}-5}{\sqrt{5}-1}\)
c. \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
d. \(\left(\sqrt{5-2\sqrt{6}+\sqrt{2}}\right).\dfrac{1}{\sqrt{3}}\)
a) \(\sqrt{\left(\sqrt{3}-3\right)^2}-\sqrt{16+6\sqrt{3}}=3-\sqrt{3}-\sqrt{\left(3+\sqrt{3}\right)^2+4}\)
b) \(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{5}-5}{\sqrt{5}-1}=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\dfrac{2\left(2-\sqrt{2}\right)}{4-2}-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5-1}}=\sqrt{5}+\sqrt{2}+2-\sqrt{2}-\sqrt{5}=2\)
c) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}=2+\sqrt{17-4\left(\sqrt{5}+2\right)}=2+\sqrt{9-4\sqrt{5}}=2+\sqrt{5}-2=\sqrt{5}\)
d) \(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\cdot\dfrac{1}{\sqrt{3}}=\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\cdot\dfrac{1}{\sqrt{3}}=1\)
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 .
Tính thu gọn :
a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)
c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)
d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\) \(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)
* Tính giá trị của biểu thức:
a. A=\(2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}\)
b. B=\(\sqrt{\left(1-\sqrt{5}\right)^2}+\sqrt{6+2\sqrt{5}}\)
c. C=\(\dfrac{1}{2-\sqrt{6}}+\dfrac{1}{2+\sqrt{6}}\)
\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=4\sqrt{2}\)
\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|1-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left|1-\sqrt{5}\right|+\left|\sqrt{5}+1\right|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
Lời giải:
a.
\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)
b.
\(B=\sqrt{(1-\sqrt{5})^2}+\sqrt{(\sqrt{5}+1)^2}=|1-\sqrt{5}|+|\sqrt{5}+1|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
c.
\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)
`a)A=2sqrt2-3sqrt{18}+4sqrt{32}-sqrt{50}`
`=2sqrt2-3sqrt{9.2}+4sqrt{16.2}-sqrt{25.2}`
`=2sqrt2-9sqrt2+16sqrt2-5sqrt2`
`=4sqrt2`
`b)B=sqrt{(1-sqrt5)^2}+sqrt{6+2sqrt5}`
`=sqrt5-1+sqrt{(sqrt5+1)^2}`
`=sqrt5-1+sqrt5+1=2sqrt5`
`c)1/(2-sqrt6)+1/(2+sqrt6)`
`=(2+sqrt6)/(4-6)+(sqrt6-2)/(6-4)`
`=(sqrt6-2-sqrt6-2)/2=-2`