Cho 2 điểm A(1;0), B(0;3). Điểm M có hoành độ dương sao cho AM=BM=5. tìm tọa độ điểm M?
Goi M co toa do la $M(x;y),x>0$
Ta co: $\overrightarrow{AM}=(x-1;y)\Rightarrow AM=\sqrt{(x-1)^2+y^2}=\sqrt{x^2+y^2-2x+1}$
$\overrightarrow{BM}=(x;y-1)\Rightarrow BM=\sqrt{x^2+y^2-2y+1}$
theo bai ra ta co:$\left\{ \begin{array}{l} AM=5\\ BM=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \sqrt{x^2+y^2-2x+1}=5\\ \sqrt{y^2+x^2-2y+1}=5 \end{array} \right.$ day la hpt dx loai 2
giai he tren duok nghiem $\left\{ \begin{array}{l} x=4\\ y=4 \end{array} \right.$
Vay $M(4;4)$
k, dap an pn dung ui do, minh tính lại ui, nhung k pk sao hệ thống nó báo sai. –  tieutulitipro 16-01-14 09:55 PM
m` cung k biet' nua~ b giai lai xem thu, cach lam` thi` la` vay do' ban :)) –  never give up 16-01-14 09:51 PM
sao đáp án đúng mk mình nhập kq vào thì nó báo sai.@@ –  tieutulitipro 16-01-14 09:51 PM
hình nhưư đáp số bị sai bạn, chac pn giải hệ pi sai ui. –  tieutulitipro 16-01-14 09:48 PM
k co' j b:)0 con` 1 cach nua~ la` b viet ptdt d la duong trung truc cua AB khi do' M thuoc d, bieu dien toa do M theo tso t ..... cach' nay` hoi dai` :P –  never give up 16-01-14 08:55 PM
tks pn nhiu nha –  tieutulitipro 16-01-14 08:53 PM

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