\(\sqrt{29}>\sqrt{25}\)= 5
\(\sqrt{3}>1\)
\(\sqrt{2003}>\sqrt{1936}=44\)
Cộng từng vế của ba bất đẳng thức ta được 
\(\sqrt{29}+\sqrt{3}+\sqrt{2003}\) > 1+5 +44 = 50

\(\sqrt{29}>\sqrt{25}=5\)

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

\(\sqrt{2003}>\sqrt{1936}=44\)

\(=>\sqrt{29}+\sqrt{3}+\sqrt{2003}>5+1+44=50\)

\(\sqrt{29}+\sqrt{3}+\sqrt{2003}>\sqrt{25}+\sqrt{1}+\sqrt{1936}=5+1+44=50\)

\(\text{Vậy }\sqrt{29}+\sqrt{3}+\sqrt{2003}>50\)

\(\sqrt{29}+\sqrt{3}+\sqrt{2013}>\sqrt{25}+\sqrt{1}+\sqrt{1936}=5+1+44=50\)

\(\sqrt{2004}-\sqrt{2003}=\dfrac{1}{\sqrt{2004}+\sqrt{2003}}\)

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

Mà \(\sqrt{2004}+\sqrt{2003}< \sqrt{2006}< \sqrt{2005}\)

\(\Rightarrow\dfrac{1}{\sqrt{2004}+\sqrt{2003}}>\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\Rightarrow\sqrt{2004}-\sqrt{2003}>\sqrt{2006}-\sqrt{2005}\)

\(\sqrt{29}+\sqrt{3}+\sqrt{2015}>\sqrt{25}+\sqrt{1}+\sqrt{1936}\)\(=5+1+44=50\)

\(\text{Vậy }\sqrt{29}+\sqrt{3}+\sqrt{2015}>50\)

50 bé hơn

đúng 100%

50 bé hơn đó bạn !!! Vì mình không biết bấm căn thức nên mình phải ghi vầy !!!

ta có

\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)

\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có  \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)

\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)

So sánh gì thế em, em nhập đủ đề vào hi

Áp dụng BĐT CAuchy-Schwarz ta có:

Đặt \(A^2=\left(\sqrt{2003}+\sqrt{2005}\right)^2\)

\(\le\left(1+1\right)\left(2003+2005\right)\)

\(=2\cdot4008=8016\)

\(\Rightarrow A^2\le8016\Rightarrow A\le2\sqrt{2004}=B\)

MÌNH LỚP 7 NHƯNG TRẢ LỜI ĐƯỢC LÈ

ÂY ... >>>>>>

BI ...========

CI <<<<<<<<<

CÂU TRẢ LỜI LÀ Â B C D E F J A T O E M S D

ÂYY

a) Ta có :\(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2}\cdot\sqrt{3}=5+2\sqrt{6}>5=\left(\sqrt{5}\right)^2\)

\(\Rightarrow\left(\sqrt{2}+\sqrt{3}\right)^2>\left(\sqrt{5}\right)^2\Leftrightarrow\sqrt{2}+\sqrt{3}>\sqrt{5}\)

a) \(\sqrt{2}+\sqrt{3}>\sqrt{5}\)

b) \(\sqrt{2003}+\sqrt{2005}< 2.\sqrt{2004}\)

HOK TOT

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

\(=2\cdot2004+2\sqrt{2004^2-1}\)

Mà \(2004^2-1< 2004^2\Rightarrow2\cdot2004+2\sqrt{2004^2-1}< 2\cdot2004+2\sqrt{2004^2}=2\cdot2004+2\cdot2004=4\cdot2004\)

Mặt khác \(\left(2\sqrt{2004}\right)^2=4\cdot2004\)

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

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

k mik nha

Đặt \(A^2=\left(\sqrt{2003}+\sqrt{2004}\right)^2>0\)

\(\le\left(1+1\right)\left(2003+2004\right)=2\cdot4007=8014\)

\(\Rightarrow A^2\le8014\). Và 

\(B^2=\left(2\sqrt{2004}\right)^2=4\cdot2004=8016\)

Suy ra \(A^2\le8014< 8016=B^2\Leftrightarrow A< B\)