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

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

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

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

\(108\)

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

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

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

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

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

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

\(=12\cdot9=108\)

2) \(-x^2+4x-2\)

\(=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)\)

\(=-\left(x-2\right)^2+2\)

Ta có: \(-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)

Vậy: GTLN của bt là 2 tại x=2

b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))

Mà: \(\sqrt{2x^2-3}\ge0\forall x\)

Dấu "=" xảy ra:

\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)

Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)

...

1:

b: \(4\sqrt{5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{75}\)

=>\(4\sqrt{5}>5\sqrt{3}\)

=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)

c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)

=>\(3-2\sqrt{5}< 1-\sqrt{5}\)

d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)

=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)

e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)

\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)

=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)

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

Ta có

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Từ đó ta có

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)

\(=1-\frac{1}{\sqrt{2005}}=\frac{\sqrt{2005}-1}{\sqrt{2005}}\)