展开 f(x)((2(x-1))/3+(3(2x+1))/9 =9)

解答

展开

f (x) (\frac{2 ( x - 1 )}{3} + \frac{3 ( 2 x + 1 )}{9} = 9)

\frac{4 x f ( x )}{3} - \frac{f ( x )}{3} = 9

求解步骤

f (x) (\frac{2 ( x - 1 )}{3} + \frac{3 ( 2 x + 1 )}{9} = 9)

= f (x) (\frac{2 ( x - 1 )}{3} + 9 = \frac{3 ( 2 x + 1 )}{9})

使用分配律:

a (b + c) = ab + a c

a = f (x), b = \frac{2 ( x - 1 )}{3}, c = \frac{3 ( 2 x + 1 )}{9} = 9

f (x) \frac{2 ( x - 1 )}{3} + f (x) \frac{3 ( 2 x + 1 )}{9} = 9

\frac{2 ( x - 1 )}{3} f (x) + \frac{3 ( 2 x + 1 )}{9} f (x) = 9

乘开

\frac{2 ( x - 1 )}{3} f (x) + \frac{3 ( 2 x + 1 )}{9} f (x) : \frac{4 x f ( x )}{3} - \frac{f ( x )}{3}

\frac{4 x f ( x )}{3} - \frac{f ( x )}{3} = 9

e x p an d - \frac{1}{x ( x + 1 )}

e x p an d (3 x - y^{2} + 1) d x

e x p an d V (- 4, - 3) \cdot P (- 7, - 6)

e x p an d \frac{t ^{2} - 1}{( t ^{2} + 1 ) ^{2}}

e x p an d Q (- 5, 3) (3, - 2)