ÅÃB−1CÓ F(X1) =POA2−X21 ⇔F(−A) =A√2⇔K2⇔=A√3A3+A32K= 3√2A2
3.Åã
B
−1Có f(x1
) =pOA2
−x2
1
⇔f(−a) =a√2⇔k2⇔=a√3a3
+a3
2k= 3√2a2
.Å1⇒f(x) = 3√.3x3
−a2
x2α2
0
ZÅ 1f(x) dx= 3√S1
== 9√8 a2
.2a2
12x4
−a2
2x2
−a
√
3
√63 =2·a√S2
=S4AM O
= 12f(−a)·M O = 12 a2
.2a√3Vậy S1
S2
= 3√