zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

solvefor x, r/f (x)=2x^2-2x-12

解答

求解

x, \frac{r}{f} (x) = 2 x^{2} - 2 x - 12

解答

x = \frac{2 f + r + 100 f ^{2} + 4 r f + r ^{2}}{4 f}, x = \frac{2 f + r - 100 f ^{2} + 4 r f + r ^{2}}{4 f}; f \neq = 0

求解步骤

\frac{r}{f} (x) = 2 x^{2} - 2 x - 12

在两边乘以

f

x r = 2 x^{2} f - 2 x f - 12 f; f \neq = 0

交换两边

2 x^{2} f - 2 x f - 12 f = x r; f \neq = 0

将

x r

para o lado esquerdo

2 x^{2} f - 2 x f - 12 f - x r = 0; f \neq = 0

改写成标准形式

a x^{2} + b x + c = 0

2 f x^{2} - (2 f + r) x - 12 f = 0; f \neq = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 2 f - r ) \pm ( - 2 f - r ) ^{2} - 4 \cdot 2 f ( - 12 f )}{2 \cdot 2 f}

化简

(- 2 f - r)^{2} - 4 \cdot 2 f (- 12 f) : 100 f^{2} + 4 r f + r^{2}

x_{1, 2} = \frac{- ( - 2 f - r ) \pm 100 f ^{2} + 4 r f + r ^{2}}{2 \cdot 2 f}; f \neq = 0

将解分隔开

x_{1} = \frac{- ( - 2 f - r ) + 100 f ^{2} + 4 r f + r ^{2}}{2 \cdot 2 f}, x_{2} = \frac{- ( - 2 f - r ) - 100 f ^{2} + 4 r f + r ^{2}}{2 \cdot 2 f}

x = \frac{- ( - 2 f - r ) + 100 f ^{2} + 4 r f + r ^{2}}{2 \cdot 2 f} : \frac{2 f + r + 100 f ^{2} + 4 r f + r ^{2}}{4 f}

x = \frac{- ( - 2 f - r ) - 100 f ^{2} + 4 r f + r ^{2}}{2 \cdot 2 f} : \frac{2 f + r - 100 f ^{2} + 4 r f + r ^{2}}{4 f}

二次方程组的解是:

x = \frac{2 f + r + 100 f ^{2} + 4 r f + r ^{2}}{4 f}, x = \frac{2 f + r - 100 f ^{2} + 4 r f + r ^{2}}{4 f}; f \neq = 0

作图

流行的例子

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

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

solvefor x,f(x^2+y^2+z^2z^2-2xy)=0

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

3 = xx + xx + xx

solvefor x,3x+4y(3x^2+4y^2)=0

so l v e f or x, 3 x + 4 y (3 x^{2} + 4 y^{2}) = 0

4x^2-2(a^4-b^4)x+a^2b^2=0

4 x^{2} - 2 (a^{4} - b^{4}) x + a^{2} b^{2} = 0