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人気のある代数 >

solvefor z,2q^2=z^2p^2(1-p^2)

解

解く

z, 2 q^{2} = z^{2} \cdot p^{2} \cdot (1 - p^{2})

解

z = \frac{2 q}{p 1 - p ^{2}}, z = - \frac{2 q}{p 1 - p ^{2}}; p \neq = 0, p \neq = - 1, p \neq = 1

解答ステップ

2 q^{2} = z^{2} p^{2} (1 - p^{2})

辺を交換する

z^{2} p^{2} (1 - p^{2}) = 2 q^{2}

以下で両辺を割る

p^{2} (1 - p^{2}); p \neq = 0, p \neq = - 1, p \neq = 1

z^{2} = \frac{2 q ^{2}}{p ^{2} ( 1 - p ^{2} )}; p \neq = 0, p \neq = - 1, p \neq = 1

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

z = \frac{2 q ^{2}}{p ^{2} ( 1 - p ^{2} )}, z = - \frac{2 q ^{2}}{p ^{2} ( 1 - p ^{2} )}; p \neq = 0, p \neq = - 1, p \neq = 1

簡素化

\frac{2 q ^{2}}{p ^{2} ( 1 - p ^{2} )} : \frac{2 q}{p 1 - p ^{2}}

簡素化

- \frac{2 q ^{2}}{p ^{2} ( 1 - p ^{2} )} : - \frac{2 q}{p 1 - p ^{2}}

z = \frac{2 q}{p 1 - p ^{2}}, z = - \frac{2 q}{p 1 - p ^{2}}; p \neq = 0, p \neq = - 1, p \neq = 1

グラフ

人気の例

(((4/3)a+5)/3)^2=100

(\frac{( \frac{4}{3} ) a + 5}{3})^{2} = 100

solvefor x,(x-3)^2+(y-6)^2=0

so l v e f or x, (x - 3)^{2} + (y - 6)^{2} = 0

x^2+2(m-1)x+(m+5)=0

x^{2} + 2 (m - 1) x + (m + 5) = 0

solvefor x,f(2x-1)=4x^2-10x+a

so l v e f or x, f (2 x - 1) = 4 x^{2} - 10 x + a

m^{2} + \frac{m}{2} - \frac{3}{2} = 0