Đáp án:
21. a, x>-1
b, min F(x)=0 , max $F(x)=5+2\sqrt{5}$
24. \(\left[ \begin{array}{l}C(1,2;1,6);D(2;2);B(-1,8;2,6)\\D(1,2;1,6);C(2;2);B(-0,2;3,4)\end{array} \right.\)
$\left \{ {{P=4\sqrt{5}} \atop {S=5}} \right.$
Giải thích các bước giải:
21. a, bpt ⇔ $x+1>0 ⇔ x>-1$
b, ĐKXĐ :$ 0≤x≤\dfrac{5}{2}$
$2F(x) = 2x+2\sqrt{2x}+1+5-2x+2\sqrt{5-2x}+1-2$
= $(\sqrt{2x}+1)²+(\sqrt{5-2x}+1)²-2≥-2$
⇒ $F(x)≥-1$
⇒ min ⇔ x=$\dfrac{5}{2}$ ⇔ F(x)=0
max ⇔ x=0 ⇔ $F(x)=5+2\sqrt{5}$
24. $d(A;Δ)=AD=\sqrt{5}$⇒$AC=\sqrt{10}$
$C∈Δ ⇒C(2t-2;t)$
$AC=\sqrt{10}⇔(t+1)²+(2t-5)²=10$
⇔ \(\left[ \begin{array}{l}t=1,6\\t=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}C(1,2;1,6);D(2;2)\\D(1,2;1,6);C(2;2)\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}I(0,1;2,3)\\I(0,5;2,5)\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}B(-1,8;2,6)\\B(-0,2;3,4)\end{array} \right.\)
⇒ $\left \{ {{P=4\sqrt{5}} \atop {S=5}} \right.$
