solvefor l,z=((1+l))/(((3+l)^2))

解答

求解

l, z = \frac{( 1 + l )}{( ( 3 + l ) ^{2} )}

l = \frac{- 6 z + 1 + - 8 z + 1}{2 z}, l = \frac{- 6 z + 1 - - 8 z + 1}{2 z}; z \neq = 0

求解步骤

z = \frac{( 1 + l )}{( ( 3 + l ) ^{2} )}

化简

\frac{( 1 + l )}{( ( 3 + l ) ^{2} )} : \frac{1 + l}{( 3 + l ) ^{2}}

z = \frac{1 + l}{( 3 + l ) ^{2}}

在两边乘以

(3 + l)^{2}

z (3 + l)^{2} = 1 + l

解

z (3 + l)^{2} = 1 + l : l = \frac{- 6 z + 1 + - 8 z + 1}{2 z}, l = \frac{- 6 z + 1 - - 8 z + 1}{2 z}; z \neq = 0

l = \frac{- 6 z + 1 + - 8 z + 1}{2 z}, l = \frac{- 6 z + 1 - - 8 z + 1}{2 z}; z \neq = 0

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}

\frac{( 5 + x )}{( 5 - x )} - \frac{( 5 - x )}{( 5 + x )} = \frac{33}{4}