1/x-1/((x+b))= 1/a-1/((a+b))

解答

\frac{1}{x} - \frac{1}{( x + b )} = \frac{1}{a} - \frac{1}{( a + b )}

x = \frac{- b + b ^{2} + 4 a ( b + a )}{2}, x = - \frac{b + b ^{2} + 4 a ( b + a )}{2}; b \neq = 0

求解步骤

\frac{1}{x} - \frac{1}{( x + b )} = \frac{1}{a} - \frac{1}{( a + b )}

乘以最小公倍数

a (x + b) (b + a) - a x (b + a) = x (x + b) (b + a) - a x (x + b)

解

a (x + b) (b + a) - a x (b + a) = x (x + b) (b + a) - a x (x + b) : x = \frac{- b + b ^{2} + 4 a ( b + a )}{2}, x = - \frac{b + b ^{2} + 4 a ( b + a )}{2}; b \neq = 0

x = \frac{- b + b ^{2} + 4 a ( b + a )}{2}, x = - \frac{b + b ^{2} + 4 a ( b + a )}{2}; b \neq = 0

so l v e f or l, z = \frac{( 1 + l )}{( ( 3 + l ) ^{2} )}

so l v e f or x, 82 f = \frac{6}{( 2 - 4 x )}

\frac{1}{2 ( 4 - x )} = 8 - 2.5 x

0.25 = \frac{1}{n}

\frac{e ^{- 4}}{t} = \frac{1}{4}