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NM
1
27 tháng 6 2017
\(\sqrt{7-2\sqrt{2+5\sqrt{2}+\sqrt{18-2\cdot4\cdot\sqrt{2}}}}\)=\(\sqrt{7-2\sqrt{2+5\sqrt{2}+4-\sqrt{2}}}\)
=\(\sqrt{7-2\sqrt{6+4\sqrt{2}}}=\sqrt{7-2\left(2+\sqrt{2}\right)}\) =\(\sqrt{3+2\sqrt{2}}\) =\(\sqrt{2}+1\)
TM
1
24 tháng 9 2019
\(\sqrt{18-\sqrt{128}}=\sqrt{18-8\sqrt{2}}=\sqrt{16-2.4.\sqrt{2}+2}=\sqrt{\left(4-\sqrt{2}\right)^2}=4-\sqrt{2}\)
=> \(\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}=\sqrt{2+5\sqrt{2}+4-\sqrt{2}}=\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{4+2.2\sqrt{2}+2}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
=> \(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{7-2\left(2+\sqrt{2}\right)}=\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}.1+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
KT
6 tháng 7 2018
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
KT
6 tháng 7 2018
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
QP
25 tháng 10 2016
1,=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3}-\sqrt{\sqrt{12}+4}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
=\(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
=\(\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{4+2\sqrt{3}}\)
=\(\sqrt{3}+1\)
Y
17 tháng 6 2019
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+\sqrt{12}}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\frac{4+\sqrt{4^2-7}}{2}}+\sqrt{\frac{4-\sqrt{4^2-7}}{2}}-\left(\sqrt{\frac{4+\sqrt{4^2-7}}{2}}-\sqrt{\frac{4-\sqrt{4^2-7}}{2}}\right)+\left(\sqrt{3}+1\right)^2\)
( áp dụng công thức căn phức tạp )
\(=2\sqrt{\frac{4-3}{2}}+4+2\sqrt{3}\)
\(=\sqrt{2}+4+2\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 6 2019
\(A=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\left(4-\sqrt{2}\right)}}}\)
\(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{4+2\sqrt{3}}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}+\left(\sqrt{3}+1\right)^2=\sqrt{2}+4+2\sqrt{3}\)
27 tháng 7 2017
Rút gọn :
\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|=\sqrt{5}-3+\sqrt{5}-2=2\sqrt{5}-5\)
Tính :
a) \(\sqrt{8,1.250}=\sqrt{81.25}=\sqrt{81}.\sqrt{25}=9.5=45\)
b) \(\sqrt{\dfrac{10.4,9}{16}}=\sqrt{\dfrac{1.49}{16}}=\dfrac{\sqrt{1}.\sqrt{49}}{\sqrt{16}}=\dfrac{7}{4}\)
c) \(\sqrt{8}.\sqrt{50}=\sqrt{8.50}=20\)
d) \(\dfrac{\sqrt{128}}{\sqrt{18}}=\sqrt{\dfrac{128}{18}}\approx2,7\)
27 tháng 7 2017
\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =3-\sqrt{5}+\sqrt{5}-2=1\)
WR
23 tháng 6 2019
a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{6-2\left(1+\sqrt{3}\right)}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=1+\sqrt{3}\)
b) Tương tự a) đ/s =5
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
\(A=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}\)
\(A=\sqrt{6-2\sqrt{4+\sqrt{12}}}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(A=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(B=\sqrt{25}=5\)
kiểm tra bằng máy tính:
\(2\sqrt{2+\sqrt{50}\sqrt{18-\sqrt{128}}}>7\)
căn thức ko có nghĩa
tại sao vậy??????