2LOG 2 2 X − 14LOG 4 X + = 3 0 ⇔ 2LOG 2 2 X − 7 LOG 2 X + = 3 0LOG 1⇔ LOG 2 X = 3 HAY 2X = 2 ⇔ X = 2 3 = 8 HAY X = 2 1 2 = 21 1X − X + X = − + =1 1 1 12 2 4 3 2( ( 2 )I = ∫ X X − DX = ∫ X − X + X DX = 5 4 3 1( )

1) - DE VA DAP AN TN MON TOAN 2010
Toán DE VA DAP AN TN MON TOAN 2010

1) 2log 2 2 x − 14log 4 x + = 3 0 ⇔ 2log 2 2 x − 7 log 2 x + = 3 0

log 1

⇔ log 2 x = 3 hay 2

x = 2 ⇔ x = 2 3 = 8 hay x = 2 1 2 = 2

1 1

x − x + x = − + =

1 1 1 1

2 2 4 3 2

( 1) ( 2 )

I = ∫ x x − dx = ∫ x − x + x dx = 5 4 3 1

( )