\(\sqrt{\left(\sqrt{7}-5\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
Tính giá trị của biểu thức
a. \(A=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
b. \(B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
c. \(C=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
d. \(D=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
a) \(A=\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)
\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}=2\sqrt{2}-2\)
b) \(B=\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)
\(=2\sqrt{2}-7+3-2\sqrt{2}=-4\)
c) \(C=\sqrt{9+6\sqrt{2}+2}-\sqrt{9-6\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=\left(3+\sqrt{2}\right)-\left(3-\sqrt{2}\right)=2\sqrt{2}\)
d) \(D=\sqrt{9+12\sqrt{2}+8}+\sqrt{9-12\sqrt{2}+8}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}=\left(3+2\sqrt{2}\right)-\left(3-2\sqrt{2}\right)=4\sqrt{2}\)
a.\(\sqrt{17}-6\sqrt{2}+3+\sqrt{2 }\)
b.\(\left(3+\sqrt{ }5\right).\left(\sqrt{ }10.\sqrt{ }2\right).\sqrt{3-\sqrt{ }5}\)
c.\(\left(\sqrt{2}-3\right).\sqrt{11+6\sqrt{2}}\)
d.\(\sqrt{23+8\sqrt{7}}-\sqrt{2}\)
nhanh nha gấp lắm trcs 9h
\(a,=\sqrt{17}-5\sqrt{2}+3\\ b,=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\\ =\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\\ =\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=8\\ c,=\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)=2-9=-7\\ d,4+\sqrt{7}-\sqrt{2}\)
Tính giá trị của biểu thức:
a)A=\(\sqrt{\left(2-\sqrt{5}\right)^2}\) +\(\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
b)B=\(\sqrt{6+2\sqrt{5}}\) - \(\sqrt{6-2\sqrt{5}}\)
c)C=\(\sqrt{17+12\sqrt{2}}\) + \(\sqrt{17-12\sqrt{2}}\)
a) A= \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
= \(\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)
= \(2\left(\sqrt{2}-1\right)\)
b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)
Ta có:
B2 = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)
= \(12-2\sqrt{36-20}\)
= \(12-8\)
= \(4\)
\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2
Vậy B = 2
c) C = \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\) (C > 0)
Ta có:
C2 = \(17+12\sqrt{2}+2\sqrt{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}+\left(17-12\sqrt{2}\right)\)
= \(34+2\sqrt{289-288}\)
= \(34+2\)
= \(36\)
\(\Rightarrow C=\pm6\) nhưng vì C > 0 nên C = 6
a : \(\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}\)
b : \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
c : \(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)
\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)
__
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)
__
\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)
Tính
a,\(\sqrt{4+\sqrt{15}}\)
b,\(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)
c,\(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}\)
a: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
1/ \(\frac{2}{3-\sqrt{7}}\sqrt{\frac{6\sqrt{2}-2\sqrt{14}}{3\sqrt{2}+\sqrt{14}}}\)
2/ \(\sqrt{6+2\sqrt{\sqrt{5}-\sqrt{13-\sqrt{48}}}}\)
3/ \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
4/ \(\frac{24}{\sqrt{7}+1}+\frac{4}{3+\sqrt{7}}-\frac{3}{\sqrt{7}+2}\left(4-\sqrt{7}\right)\)
5/ \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
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,\(\dfrac{6}{\sqrt{5}-2}-\dfrac{6}{\sqrt{5}+2}+\sqrt{8}-4\sqrt{\dfrac{1}{7}}\) ; \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)
Rút gọn :
\(A=\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{5}}}}\)
\(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(D=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(E=\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
A = \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
A = \(\dfrac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{5}+\sqrt{2}\right)}\)
A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\) = \(\dfrac{3}{1}\) = \(3\)
C = \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
C = \(\left(4+\sqrt{15}\right).\left(\sqrt{40-10\sqrt{15}}-\sqrt{24-6\sqrt{15}}\right)\)
C = \(\left(4+\sqrt{15}\right)\left(\sqrt{\left(5-\sqrt{15}\right)^2}-\sqrt{\left(\sqrt{15}-3\right)^2}\right)\)
C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\left(\sqrt{15}-3\right)\right)\)
C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\sqrt{15}+3\right)\)
C = \(\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
C = \(32-8\sqrt{15}+8\sqrt{15}-30=2\)
D = \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
D = \(\left(\sqrt{30-10\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\left(3+\sqrt{5}\right)\)
D = \(\left(\sqrt{\left(5-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)\left(3+\sqrt{5}\right)\)
D = \(\left(5-\sqrt{5}-\left(\sqrt{5}-1\right)\right)\left(3+\sqrt{5}\right)\)
D = \(\left(5-\sqrt{5}-\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\)
D = \(\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
D = \(18+6\sqrt{5}-6\sqrt{5}-10=8\)
E = \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{5}}\)
E = \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
E = \(3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6-1}\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)