a) $2\sqrt{6},\sqrt{29},4\sqrt{2},3\sqrt{5};$

b) $\sqrt{38},2\sqrt{14},3\sqrt{7},6\sqrt{2}$

a) \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)

b) \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)

99999999999999999999 x 100 = 9999999999999999999900

ok chưa bn

99999999999999999999x100=9999999999999999999900

a, \(ab+b\sqrt{a}+\sqrt{a}+1=\sqrt{a}b\left(\sqrt{a}+1\right)+\sqrt{a}+1\)

\(=\left(b\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\)

b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

\(=\sqrt{x^2}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y^2}\left(\sqrt{y}+\sqrt{x}\right)=\left(\left|x\right|-\left|y\right|\right)\left(\sqrt{x}+\sqrt{y}\right)\)

a) $(\sqrt{a}+1)(b\sqrt{a}+1)$.
b) $(x-y)(\sqrt{x}+\sqrt{y})$

a) $(\sqrt{a}+1)(b\sqrt{a}+1)$.
b) $(x-y)(\sqrt{x}+\sqrt{y})$.

\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)

\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)

\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)

\(\hept{\begin{cases}\frac{1}{x-2}=0,2\\\frac{1}{y-2}=0,2\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x-2}=\frac{1}{5}\\\frac{1}{y-2}=\frac{1}{5}\end{cases}}}\)ĐK : \(x;y\ne2\)

\(\Leftrightarrow\hept{\begin{cases}x-2=5\left(1\right)\\y-2=5\left(2\right)\end{cases}}\)

Từ phương trình (1) \(\Rightarrow x=5+2=7\)( tmđk )

Từ phương trinh (2) \(\Rightarrow y=5+2=7\)( tmđk )

Vậy hệ phương trình có một nghiệm ( x ; y ) = ( 7 ; 7 )