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) \(\sqrt{6,8^2-3.2^2}\)

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

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

b) \(\sqrt{21,8^2-18,2^2}\)

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

\(=\sqrt{3,6.40}=\sqrt{4.36}=2.6=12\)

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

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

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

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

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

\(=12\cdot9=108\)

giúp mk với

a) \(\sqrt{117.5^2-26.5^2-1440}\)

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

\(=5\sqrt{177-26-1440}\)

\(=5\sqrt{-1289}\)

\(=-5\sqrt{1289}\)

Câu B tương tự . 

E ms lên lớp 9 sai thì thôi nha