A09

vẽ tam giác đều ADK(K và B cùng phía với AD)

=>$\widehat{DAK }$=$60^{\circ}$=>$\widehat{KAB}$=$90^{\circ}$-$60^{\circ}=30^{\circ}$.

$\Delta ABK$ cân tại A=>$\widehat{ABK}=75^{\circ}$=>KBC=$90^{\circ}-75^{\circ}=15^{\circ}$

tương tự 

$\Delta DKC$cân tại D=>$\widehat{DKC}=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}$=>$\widehat{KCB}=15^{\circ}$

có $\Delta AEB=\Delta BKC$(g.c.g)=>AE=BK=KC

$\Delta ADE=\Delta KDC$(c.g.c)=>DE=DC(1), $\widehat{ADE}=\widehat{KDC}=30^{\circ}$=>$\widehat{EDC}=60^{\circ}$ (2)

$\overset{(1),(2)}{\rightarrow}$$\Delta EDC$ đều

206) tìm max: $\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x (x\geq \frac{-1}{2})$

bạn ơi mình giải với x$\geq 1$ nhé

$Có: x^3+3x^2-4=x^3-x^2+4x^2-4=x^2(x-1)+4(x^2-1)=(x-1)(x^2+4x+4)=(x-1)(x+2)^2)$

$x^3-3x^2+4=(x+1)(x-2)^2$

P=$\frac{(x-1)(x+2)^2+(x-2)(x+2)\sqrt{(x-1)(x+1)}}{(x+1)(x-2)^2+(x-2)(x+2)\sqrt{(x-1)(x+1)}}$

  =$\frac{(x+2)\sqrt{x-1}[(x+2)\sqrt{x-1}+\sqrt{x+1}(x-2)]}{(x-2)\sqrt{x+1}[\sqrt{x+1}(x-2)+(x+2)\sqrt{x-1}]}$

  =$\frac{(x+2)\sqrt{x-1}}{(x-2)\sqrt{x+1}}$

Bài 1:

$A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$

$=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}} =\frac{2+\sqrt{3}}{2+\sqrt{3+2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{2-\sqrt{3-2\sqrt{3}+1}} =\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}} =\frac{2+\sqrt{3}}{2+\left | \sqrt{3}+1 \right |}+\frac{2-\sqrt{3}}{2-\left | \sqrt{3}-1 \right |} =\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}} =\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{9-3}$$=\frac{6}{6}=1$

bạn ơi hình như đề bài bài 1 là: $\frac{1+\tfrac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\tfrac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$ mới đúng thì phải.