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受欢迎的代数 >

solvefor x,f(a+h)=3x^2+2x

解答

求解

x, f (a + h) = 3 x^{2} + 2 x

解答

x = \frac{- 1 + 3 f ( h + a ) + 1}{3}, x = - \frac{3 f ( h + a ) + 1 + 1}{3}

求解步骤

f (a + h) = 3 x^{2} + 2 x

展开

f (a + h) : a f + h f

a f + h f = 3 x^{2} + 2 x

交换两边

3 x^{2} + 2 x = a f + h f

将

h f

para o lado esquerdo

3 x^{2} + 2 x - h f = a f

将

a f

para o lado esquerdo

3 x^{2} + 2 x - h f - a f = 0

使用求根公式求解

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 ( - h f - a f )}{2 \cdot 3}

化简

2^{2} - 4 \cdot 3 (- h f - a f) : 2 1 + 3 f (a + h)

x_{1, 2} = \frac{- 2 \pm 2 1 + 3 f ( a + h )}{2 \cdot 3}

将解分隔开

x_{1} = \frac{- 2 + 2 1 + 3 f ( a + h )}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 1 + 3 f ( a + h )}{2 \cdot 3}

x = \frac{- 2 + 2 1 + 3 f ( a + h )}{2 \cdot 3} : \frac{- 1 + 3 f ( h + a ) + 1}{3}

x = \frac{- 2 - 2 1 + 3 f ( a + h )}{2 \cdot 3} : - \frac{3 f ( h + a ) + 1 + 1}{3}

二次方程组的解是:

x = \frac{- 1 + 3 f ( h + a ) + 1}{3}, x = - \frac{3 f ( h + a ) + 1 + 1}{3}

作图

流行的例子

(x^2+9x+20)/(x+1)>0

\frac{x ^{2} + 9 x + 20}{x + 1} > 0

化简 (x^5y^4)/(x^{-5)y^4}

s im pl i f y \frac{x ^{5} y ^{4}}{x ^{- 5} y ^{4}}

(5x-1)(2x+1)-10x^2=0.6

(5 x - 1) (2 x + 1) - 10 x^{2} = 0.6

- 7 < 4 x - 3 < 4

因式分解 x^2+12xy+32y^2

f a c t or x^{2} + 12 x y + 32 y^{2}