270= n/(2[2x^{10)+(n-1)5]}

解答

270 = \frac{n}{2 [ 2 x ^{10} + ( n - 1 ) 5 ]}

x = 10 \frac{- 2699 n + 2700}{1080}, x = - 10 \frac{- 2699 n + 2700}{1080}

求解步骤

270 = \frac{n}{2 [ 2 x ^{10} + ( n - 1 ) \cdot 5 ]}

在两边乘以

[2 x^{10} + (n - 1) 5]

270 [2 x^{10} + (n - 1) \cdot 5] = \frac{n}{2}

解

270 [2 x^{10} + (n - 1) \cdot 5] = \frac{n}{2} : x = 10 \frac{- 2699 n + 2700}{1080}, x = - 10 \frac{- 2699 n + 2700}{1080}

x = 10 \frac{- 2699 n + 2700}{1080}, x = - 10 \frac{- 2699 n + 2700}{1080}

\frac{7}{5 \cdot ( - 7 x ^{2} + 2 ) ^{2}} = \frac{5}{7}

\frac{2 x}{3 x + 6} = \frac{3}{2}

so l v e f or x, f = \frac{{ x }}{( 1 + { x } )}

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

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