u' = -3t²/2 + 3t是怎么来的,3^x+3^y=9^x+9^y3^x+3^y=(3^x)²+(3^y)²设 3^x = a ; 3^y = b;a > 0 ; b > 0; ---------(1)a + b = a² + b² -------(2)u=27^x+27^y= (3^x)³ + (3^y)³= a³ + b³= (a + b)(a² - ab +

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u' = -3t²/2 + 3t是怎么来的,3^x+3^y=9^x+9^y3^x+3^y=(3^x)²+(3^y)²设 3^x = a ; 3^y = b;a > 0 ; b > 0; ---------(1)a + b = a² + b² -------(2)u=27^x+27^y= (3^x)³ + (3^y)³= a³ + b³= (a + b)(a² - ab +
u' = -3t²/2 + 3t是怎么来的,
3^x+3^y=9^x+9^y
3^x+3^y=(3^x)²+(3^y)²
设 3^x = a ; 3^y = b;
a > 0 ; b > 0; ---------(1)
a + b = a² + b² -------(2)
u=27^x+27^y
= (3^x)³ + (3^y)³
= a³ + b³
= (a + b)(a² - ab + b²)
结合(1)(2):
u = (a + b)(a² - ab + b²)
= (a + b)(a² + b² - ab)
= (a + b)(a + b - ab)
∵(a + b)² = a² + b² + 2ab
∴ab = [(a + b)² - (a² + b²)]/2
ab = [(a+b)² - (a+b)]/2
u = (a + b)(a + b - ab)
= (a + b)(a + b - [(a+b)² - (a+b)]/2)
设 t = a + b; t >0
u = t [t - (t² - t)/2]
= t² - (t³ - t²)/2
= -t³/2 + 3t²/2 --------(t > 0)
u' = -3t²/2 + 3t
u' = 0时,取极值
t = 2 或 t = 0(排除)
∵u' > 0时,
-3t²/2 + 3t > 0
0 < t < 2;
u在此范围单调递增.
∵u' < 0时,
-3t²/2 + 3t < 0
t >2 或 t < 0(排除)
u在此范围单调递减.
综上,
u在 t = 2时取到唯一的极大值,所以t = 2时,u取最大值
t = 2 ; u = 2
0 < u ≤ 2