So sánh các căn bậc hai.
a, Giả sử 8 > $\sqrt{15}+\sqrt{17}$
Ta có (1) <=> $8^{2}>(\sqrt{15}+\sqrt{17})^{2}$ = 15 + 17 + 2$\sqrt{255}$
<=> 16 > $\sqrt{255}$ <=> 256 > 255 (2)
Ta thấy (2) đúng mà (2) <=> (1). Vậy (1) đúng hay 8 > $\sqrt{15}+\sqrt{17}$
b, Ta có: $(\sqrt{10}+\sqrt{13})^{2}$ = 10 + 13 + 2$\sqrt{13.10}$ = 23 + 2$\sqrt{130}$
$(\sqrt{11}+\sqrt{12})^{2}$ = 11 + 12 + 2$\sqrt{11.12}$ = 23 + 2$\sqrt{132}$
Vì 130 < 132 => $\sqrt{130}$ < $\sqrt{132}$
<=> 2$\sqrt{130}$ < 2$\sqrt{132}$ <=> 23 + 2$\sqrt{130}$ < 23 + 2$\sqrt{132}$
<=> $(\sqrt{10}+\sqrt{13})^{2}$ < $(\sqrt{11}+\sqrt{12})^{2}$
<=> $\sqrt{10}+\sqrt{13}$ < $\sqrt{11}+\sqrt{12}$
c, $(\sqrt{100}+\sqrt{200})^{2}$ = 100 + 200 + 2$\sqrt{100.200}$ = 300 + 2$\sqrt{20000}$
$(\sqrt{104}+\sqrt{196})^{2}$ = 104 + 196 + 2$\sqrt{104.196}$ = 300 + 2$\sqrt{20384}$
Vì 20000 < 20384 => $\sqrt{20000}$ < $\sqrt{20384}$
<=> 2$\sqrt{20000}$ < 2$\sqrt{20384}$ <=> 300 + 2$\sqrt{20000}$ < 300 + 2$\sqrt{20384}$
<=> $(\sqrt{100}+\sqrt{200})^{2}$ < $(\sqrt{104}+\sqrt{196})^{2}$
<=> $\sqrt{100}+\sqrt{200}$ < $\sqrt{104}+\sqrt{196}$
d, $(\sqrt{a}+\sqrt{a+7})^{2}$ = a + a + 7 + 2$\sqrt{a.(a+7)}$ = 2a + 7 + 2$\sqrt{a^{2}+7a}$
$(\sqrt{a+2}+\sqrt{a+5})^{2}$ = a + 2 + a + 7 + 2$\sqrt{(a+2).(a+5)}$ = 2a + 7 + 2$\sqrt{a^{2}+7a+10}$
Vì $a^{2}+7a$ < $a^{2}+7a+10$ => $\sqrt{a^{2}+7a}$ < $\sqrt{a^{2}+7a+10}$
<=> 2$\sqrt{a^{2}+7a}$ < 2$\sqrt{a^{2}+7a+10}$
<=> 2a + 7 + 2$\sqrt{a^{2}+7a}$ < 2a + 7 + 2$\sqrt{a^{2}+7a+10}$
<=> $(\sqrt{a}+\sqrt{a+7})^{2}$ < $(\sqrt{a+2}+\sqrt{a+5})^{2}$
<=> $\sqrt{a}+\sqrt{7}$ < $\sqrt{a+2}+\sqrt{a+5}$