Giải câu 3 trang 24 sách phát triển năng lực toán 9 tập 1

a, $4\sqrt{7}$ và $\sqrt{105}$

$4\sqrt{7}$ = $\sqrt{4^{2}.7}=\sqrt{112}$ 

Vì 112 > 105 => $\sqrt{112}$ > $\sqrt{105}$ => $4\sqrt{7}$ > $\sqrt{105}$

b, 11 và $3\sqrt{13}$

$11 = \sqrt{11^{2}}=\sqrt{121}$; $3\sqrt{13}$ = $\sqrt{3^{2}.13}=\sqrt{117}$ 

Vì 121 > 117 => $\sqrt{121}$ > $\sqrt{117}$ => 11 > $3\sqrt{13}$

c. $\frac{1}{2}\sqrt{61}$ và $\frac{1}{3}\sqrt{137}$

$\frac{1}{2}\sqrt{61}=\sqrt{\frac{61}{2^{2}}}=\sqrt{\frac{61}{4}}=\sqrt{\frac{549}{36}}$,

$\frac{1}{3}\sqrt{137}=\sqrt{\frac{137}{3^{2}}}=\sqrt{\frac{137}{9}}=\sqrt{\frac{548}{36}}$,

Vì $\frac{549}{36}$ > $\frac{548}{36}$ => $\sqrt{\frac{549}{36}}$ > $\sqrt{\frac{548}{36}}$ => $\frac{1}{2}\sqrt{61}$ > $\frac{1}{3}\sqrt{137}$

d, $\sqrt{15}-\sqrt{14}$ và $\sqrt{14}-\sqrt{13}$

$\sqrt{15}-\sqrt{14}=\frac{(\sqrt{15}-\sqrt{14}).(\sqrt{15}+\sqrt{14})}{\sqrt{15}+\sqrt{14}}=\frac{15-14}{\sqrt{15}+\sqrt{14}}=\frac{1}{\sqrt{15}+\sqrt{14}}$

$\sqrt{14}-\sqrt{13}=\frac{(\sqrt{14}-\sqrt{13}).(\sqrt{14}+\sqrt{13})}{\sqrt{14}+\sqrt{13}}=\frac{14-13}{\sqrt{14}+\sqrt{13}}=\frac{1}{\sqrt{14}+\sqrt{13}}$

Vì 15 > 13 => $\sqrt{15} > \sqrt{13}$ => $\sqrt{15}+\sqrt{14}$ > $\sqrt{14}+\sqrt{13}$ => $\frac{1}{\sqrt{15}+\sqrt{14}}$ < $\frac{1}{\sqrt{14}+\sqrt{13}}$

=> $\sqrt{15}-\sqrt{14}$ < $\sqrt{14}-\sqrt{13}$