\(5+2\sqrt{21}=\left(3+2\sqrt{3.7}+7\right)-5\)=\(\left(\sqrt{3}+\sqrt{7}\right)^2-5=\left(\sqrt{3}+\sqrt{7}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{7}+\sqrt{5}\right)\)

Do đó \(\frac{5+2\sqrt{21}}{\sqrt{7}+\sqrt{3}-\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{3}\sqrt{5}\right)}{\sqrt{7}+\sqrt{3}-\sqrt{5}}\)\(=\sqrt{3}+\sqrt{7}+\sqrt{5}\)
 

help me

=10( (1-√4)/(1-4) + (√4-√7)/(4-7)+.....+(√97-√100)/(97-100) )

=10 (1-100)/3

=-990/3 = -330

Mik cx l9 

k hay ko tùy bn

\(2016^3-2016=2016.\left(2016^2-1\right)\)

\(=2016.\left(2016-1\right).\left(2016+1\right)\)

\(=2017.2016.2015⋮2017\) ( đpcm )

2016³-2016=2016(2016²-1)=2016.(2016-1)(2016+1)=2015.2016.2017 chia hết cho 2017

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(A=\)\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\)\(\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(\Rightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=-\frac{1}{7}\Rightarrow-7\left(-5\sqrt{x}+2\right)=\sqrt{x}+3\)

\(\Rightarrow35\sqrt{x}-14=\sqrt{x}+3\)

\(\Rightarrow34\sqrt{x}=17\)

\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\left(tm\right)\)

Vậy với \(x=\frac{1}{4}\)thì \(A=-\frac{1}{7}\)

a) \(\sqrt{x}\)< \(\sqrt{2x-1}\)

x < 2x - 1

x - 2x < -1

-x < -1

x > 1

b) \(\sqrt{x}\le\sqrt{x+1}\)

x < x + 1

0 < 1

không có x tm