Cho x,y là các số thực thay đổi .Tìm giá trị nhỏ nhất của biểu thức :

           $A=\sqrt{(x-1)^{2} +y^{2}}+\sqrt{(x+1)^{2}+y^{2}}+\left| y-2 \right|$
http://thayquocvuong.com/uploads/news/de-thi/2006/da-toan-b-dh2006-03.jpg
câu 2 phần IV

    $A=\sqrt{(1-x)^{2}+y^{2}}+\sqrt{(x+1)^{2}+y^{2}}+\left| y-2 \right|\geq \sqrt{(1-x+x+1)^{2}+(y+y)^{2}}+\left| y-2 \right|$

    vậy $A\geq \sqrt{4+4y^{2}}+\left| y-2\right|$.Xét hàm số $f(y)=2\sqrt{1+y^{2}}+\left| y-2 \right|$

     TH1:$y\geq 2=>f(y)\geq 2\sqrt{5}$

     TH2:$y<2=>f(y)=2\sqrt{1+y^{2}}+\left| y-2\right|$

     $f'(y)=\frac{2y}{\sqrt{1+y^{2}}}-1,f'(y)=0<=>y=\frac{1}{\sqrt{3}}$

    từ $f(y)=>\min f(y)=f(\frac{1}{\sqrt{3}}=2+\sqrt{3}$ (vơi min từ âm vô cùng đến 2)

      ta có $2+\sqrt{3}<2\sqrt{5}=>mìn f(y)=2+\sqrt{3}$(với min trên toàn tập R). Ta có $A\geq f(y)\geq 2+\sqrt{3}$.Đẳng thức xảy ra khi và chỉ khi x=0,y=$\frac{1}{\sqrt{3}}$.Vậy $min A=2+\sqrt{3}$.

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