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LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0
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C©u IVa.
1
) Gäi (x ,y ),(x ,y ) lµ täa ®é c¸c ®iÓm A, B ; gäi I = (x ,y ) lµ trung ®iÓm cña ®o¹n AB ta cã :
A A B B 1 1
1
2
1
2
2
2
2
2
B
y = x , y = x , x = (x + x ) , y = (x + x ) .
A
A
B
B
1
A
B
1
A
Theo gi¶ thiÕt :
2
2
2
2 2
B
AB = 2 ⇒ AB = (x − x ) + (x − x ) = 4 .
A
B
A
2
2
2
⇒
4 = (x − x ) + (x − x ) (x + x ) =
A
B
A
B
A
B
2
2
2
2
1
=
(x − x ) [1+ (x + x ) ] = [4x − 4x x ][1+ 4x ]
A
B
A
B
1
A B
4
2
⇒
4x − 4x x =
⇒ −4x x =
2
+ 4x1
1
A B
A B
1
4
2
2
− 2x12
=
− 4x ⇒ −2x x =
1
A B
2
1+ 4x12
1
+ 4x1
MÆt kh¸c
VËy
1
2
1
2
1
1
2
2
2
2
2
1
y = (x + x ) = [(x − x ) + 2x x ]= [4x − 2x x ].
1
A
B
A
B
A B
A B
1
2
2
2
2
2
2
1
y = [4x +
− 2x ] = x +
+ 4x1
1
1
1
1
1+ 4x12
Do ®ã tËp hîp trung ®iÓm I cña AB lµ ®−êng cã ph−¬ng tr×nh
1
+ 4x
2
y = x +
2
1
2
) Kh«ng gi¶m tÝnh tæng qu¸t ta cã thÓ gi¶ thiÕt r»ng x < x .Khi ®ã ta thÊy diÖn tÝch phÇn mÆt ph¼ng bÞ giíi h¹n bëi
A B
parabol vµ c¸t tuyÕn AB chÝnh lµ :
xA
1
2
S = (x − x )(x + x ) − x2dx =
2
2
B
B
A
A
∫
x
A
1
2
1
3
2 2 3
(x − x )(x + x ) − [x − x ] =
B A A B B
3
=
=
A
2
2
2
x + x + x x
B B A
3
2
=
x + x
B
A
A
(x − x )
−
B
A
2
1
6
3
=
(x − x )
B A
Râ rµng | x − x | ≤ AB = 2, ®¼ng thøc x¶y ra
B
A
⇔
AB// = x x ⇔ x = −1 , xB =1,
A B A
1
6
4
3
nªn S ≤ .8 = , ®¼ng thøc x¶y ra ⇔ x = −1,x =1.
A
B
C©u IVb.
1
) Gäi I, J lÇn l−ît lµ trung ®iÓm cña AB vµ CD, OK ⊥ AD.
Tam gi¸c AOD vu«ng ë O. Do ®ã :