ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある代数 >

solvefor x, 9/(x^2)=+q/((x+2)^2)

解

解く

x, \frac{9}{x ^{2}} = + \frac{q}{( x + 2 ) ^{2}}

解

x = \frac{6 ( q - 3 )}{9 - q}, x = - \frac{6 ( q + 3 )}{- q + 9}; q \neq = 9

解答ステップ

\frac{9}{x ^{2}} = \frac{q}{( x + 2 ) ^{2}}

たすき掛け

9 (x + 2)^{2} = x^{2} q

解く

9 (x + 2)^{2} = x^{2} q : x = \frac{6 ( q - 3 )}{9 - q}, x = - \frac{6 ( q + 3 )}{- q + 9}; q \neq = 9

x = \frac{6 ( q - 3 )}{9 - q}, x = - \frac{6 ( q + 3 )}{- q + 9}; q \neq = 9

グラフ

人気の例

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

\frac{x ^{2} - x - 1}{( x - 1 )} = \frac{1 - x}{( x - 1 )}

(j)*((x-2))/((x+2))+((x+2))/((x-2))=4

(j) \cdot \frac{( x - 2 )}{( x + 2 )} + \frac{( x + 2 )}{( x - 2 )} = 4

x+1/x =3,((5x))/((4x^2-5x+4))

x + \frac{1}{x} = 3, \frac{( 5 x )}{( 4 x ^{2} - 5 x + 4 )}

1/v = 1/(1.5)+1/5

\frac{1}{v} = \frac{1}{1.5} + \frac{1}{5}

solvefor t,c= s/t 1+s/(t^2)

so l v e f or t, c = \frac{s}{t} 1 + \frac{s}{t ^{2}}