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

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

bạn lê nhat phuong oi sai rồi

câu đc  của bạn phương thiếu căn ở 1440

câu a đc 108

câu b đc128

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

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

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

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

\(108\)

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

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

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

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

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

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

\(=12\cdot9=108\)