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人気のある代数 >

solvefor a,(a+b+c+d)^2=4(a+b)(c+d)

解

解く

a, (a + b + c + d)^{2} = 4 (a + b) (c + d)

解

a = - b + c + d

解答ステップ

(a + b + c + d)^{2} = 4 (a + b) (c + d)

拡張

(a + b + c + d)^{2} : 2 ab + 2 b c + 2 b d + b^{2} + a^{2} + 2 a c + c^{2} + 2 a d + 2 c d + d^{2}

拡張

4 (a + b) (c + d) : 4 a c + 4 a d + 4 b c + 4 b d

2 ab + 2 b c + 2 b d + b^{2} + a^{2} + 2 a c + c^{2} + 2 a d + 2 c d + d^{2} = 4 a c + 4 a d + 4 b c + 4 b d

両辺から

4 b c + 4 b d

を引く

2 ab + 2 b c + 2 b d + b^{2} + a^{2} + 2 a c + c^{2} + 2 a d + 2 c d + d^{2} - (4 b c + 4 b d) = 4 a c + 4 a d + 4 b c + 4 b d - (4 b c + 4 b d)

簡素化

b^{2} + 2 ab - 2 b c - 2 b d + a^{2} + c^{2} + 2 a c + d^{2} + 2 a d + 2 c d = 4 a c + 4 a d

両辺から

4 a c + 4 a d

を引く

b^{2} + 2 ab - 2 b c - 2 b d + a^{2} + c^{2} + 2 a c + d^{2} + 2 a d + 2 c d - (4 a c + 4 a d) = 4 a c + 4 a d - (4 a c + 4 a d)

簡素化

a^{2} + (2 b - 2 c - 2 d) a + b^{2} + c^{2} - 2 b c + d^{2} - 2 b d + 2 c d = 0

解くとthe二次式

a_{1, 2} = \frac{- ( 2 b - 2 c - 2 d ) \pm ( 2 b - 2 c - 2 d ) ^{2} - 4 \cdot 1 \cdot ( b ^{2} + c ^{2} - 2 b c + d ^{2} - 2 b d + 2 c d )}{2 \cdot 1}

(2 b - 2 c - 2 d)^{2} - 4 \cdot 1 \cdot (b^{2} + c^{2} - 2 b c + d^{2} - 2 b d + 2 c d) = 0

a_{1, 2} = \frac{- ( 2 b - 2 c - 2 d ) \pm 0}{2 \cdot 1}

a = \frac{- ( 2 b - 2 c - 2 d )}{2 \cdot 1}

簡素化

\frac{- ( 2 b - 2 c - 2 d )}{2 \cdot 1} : - b + c + d

a = - b + c + d

二次equationの解:

a = - b + c + d

グラフ

人気の例

x^{2} - 4 x - 24 = - 9 x

x^{2} - 3 x - 38 = - 10

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

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

x^{2} + 12 x - 253 = 0

(2x-1)^2-(x+1)^2=9

(2 x - 1)^{2} - (x + 1)^{2} = 9