a) \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)

\(=\sqrt{91.144-1440}=\sqrt{144\left(91-10\right)}=\sqrt{12^2.9^2}=12.9=108\)

b) \(\sqrt{146,5^2-109,5^2+27.256}=\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)

\(=\sqrt{37.256+27.256}=\sqrt{256\left(37+27\right)}=\sqrt{256.64}=\sqrt{16^2.8^2}=16.8=128\)

\(\sqrt{117,5^2-26,5^2}-1440=-202475\)

\(\sqrt{146,5^2-109,5^2+27,256=}-11816494\) 

bạn lê nhat phuong oi sai rồi

câu đc  của bạn phương thiếu căn ở 1440

câu a đc 108

câu b đc128

2)

\(=4008+2\sqrt{\left(2004-1\right)\left(2004+1\right)}=4008+2\sqrt{2004^2-1}\)

\(=4008+2\sqrt{2004^2}\)

Ta có \(2004^2>2004^2-1\Rightarrow\sqrt{2004^2}>\sqrt{2004^2-1}\Rightarrow4008+2\sqrt{2004^2}>4008+2\sqrt{2004^2-1}\)

Vậy \(2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)

1.  a) 108
     b) 128
2.  >

cái mình cần là cách giải bạn ạ

\(a=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\left(10\right)}=\sqrt{36}=6\)

a) \(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}\)

=\(\sqrt{3,6.10}=\sqrt{36}=6\)

b)\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\)

=\(\sqrt{3,6.40}=\sqrt{144}=12\)

c)\(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)

=\(\sqrt{91.144-1440}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=108\)

d)\(\sqrt{146,5^2-109,5^2+27.256}\)=\(\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)

=\(\sqrt{37.256+\sqrt{27.256}}=\sqrt{64.256}=\sqrt{64}.\sqrt{256}=128\)

c) \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-144.10}\)

\(=\sqrt{144.91-144.10}=\sqrt{144.\left(91-10\right)}=12\sqrt{81}=12.9=108\)

\(\sqrt{117.5^2-26.5^2-1440}\)

\(=\sqrt{\left(117.5-26.5\right)\left(117.5+26.5\right)-1440}\)

\(=\sqrt{91\cdot144-1440}\)

\(=12\cdot9=108\)

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)

\(=\sqrt{3}-1\)

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

\(=3-2\sqrt{2}+3\sqrt{2}+1\)

\(=4+\sqrt{2}\)

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

\(=2\sqrt{2}-2+2\sqrt{2}+1\)

\(=4\sqrt{2}-1\)

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)

\(108\)

\(=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\\ =\sqrt{91\cdot144-1440}=\sqrt{144\left(91-10\right)}\\ =\sqrt{144\cdot81}=\sqrt{144}\cdot\sqrt{81}=12\cdot9=108\)

\(K=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{81-17}-8=\sqrt{64}-8=8-8=0\)

\(=\sqrt{81-17}-8\)

=8-8

=0