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受欢迎的代数 >

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

解答

求解

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

解答

x = \frac{f + f ^{2} - 4 ( h - h f )}{2}, x = \frac{f - f ^{2} - 4 ( h - h f )}{2}

求解步骤

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

展开

f (x + h) : x f + h f

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

交换两边

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

将

h f

para o lado esquerdo

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

将

x f

para o lado esquerdo

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

改写成标准形式

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

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

使用求根公式求解

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

化简

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

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

将解分隔开

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

x = \frac{- ( - f ) + f ^{2} - 4 ( h - h f )}{2 \cdot 1} : \frac{f + f ^{2} - 4 ( h - h f )}{2}

x = \frac{- ( - f ) - f ^{2} - 4 ( h - h f )}{2 \cdot 1} : \frac{f - f ^{2} - 4 ( h - h f )}{2}

二次方程组的解是:

x = \frac{f + f ^{2} - 4 ( h - h f )}{2}, x = \frac{f - f ^{2} - 4 ( h - h f )}{2}

作图

流行的例子

-332.65=(0.0004)x^2-0.7281x

- 332.65 = (0.0004) x^{2} - 0.7281 x

0.11 = (0.5 + x) x

(x+(x^2+4))/5 =2

\frac{x + ( x ^{2} + 4 )}{5} = 2

24 t - 0.48 t^{2} = 0

3/4 x[ 2/7+5/8 ]= 3/4 (x^2)/7

\frac{3}{4} x [\frac{2}{7} + \frac{5}{8}] = \frac{3}{4} \frac{x ^{2}}{7}