x^2+x= 2/(x^2+x+1)

解答

x^{2} + x = \frac{2}{x ^{2} + x + 1}

x = \frac{5 - 1}{2}, x = - \frac{5 + 1}{2}

十进制

x = 0.61803 \dots, x = - 1.61803 \dots

求解步骤

x^{2} + x = \frac{2}{x ^{2} + x + 1}

在两边乘以

x^{2} + x + 1

x^{2} (x^{2} + x + 1) + x (x^{2} + x + 1) = 2

解

x^{2} (x^{2} + x + 1) + x (x^{2} + x + 1) = 2 : x = \frac{5 - 1}{2}, x = - \frac{5 + 1}{2}

x = \frac{5 - 1}{2}, x = - \frac{5 + 1}{2}

\frac{( 1.2 - n )}{( 1 - n )} = 1.4

\frac{( x - 3 )}{x} = \frac{( x + 3 )}{( x + 8 )}

\frac{2 x + 4}{5 x} = 23 x - 6

\frac{( 0.5 ( x ^{5} ) )}{( 5 x ^{3} - 2 x ^{2} + 3 x + 7 )} = 5

(\frac{5 x}{4 x}) + 2 x = 6