So sánh các căn bậc hai

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$\sqrt{a^2+7a+10}$

<=> $(\sqrt{a}+\sqrt{a+7}{)}^2$ < $(\sqrt{a+2}+\sqrt{a+5}{)}^2$

<=> $\sqrt{a}+\sqrt{7}$ < $\sqrt{a+2}+\sqrt{a+5}$