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受欢迎的代数 >

1+4+7+10(3n-2)=(n(3n-1))/2

解答

1 + 4 + 7 + 10 (3 n - 2) = \frac{n ( 3 n - 1 )}{2}

解答

n = \frac{61 + 3529}{6}, n = \frac{61 - 3529}{6}

+1

十进制

n = 20.06756 \dots, n = 0.26576 \dots

求解步骤

1 + 4 + 7 + 10 (3 n - 2) = \frac{n ( 3 n - 1 )}{2}

在两边乘以

2

2 + 8 + 14 + 20 (3 n - 2) = n (3 n - 1)

展开

2 + 8 + 14 + 20 (3 n - 2) : 60 n - 16

展开

n (3 n - 1) : 3 n^{2} - n

60 n - 16 = 3 n^{2} - n

交换两边

3 n^{2} - n = 60 n - 16

将

16

para o lado esquerdo

3 n^{2} - n + 16 = 60 n

将

60 n

para o lado esquerdo

3 n^{2} - 61 n + 16 = 0

使用求根公式求解

n_{1, 2} = \frac{- ( - 61 ) \pm ( - 61 ) ^{2} - 4 \cdot 3 \cdot 16}{2 \cdot 3}

(- 61)^{2} - 4 \cdot 3 \cdot 16 = 3529

n_{1, 2} = \frac{- ( - 61 ) \pm 3529}{2 \cdot 3}

将解分隔开

n_{1} = \frac{- ( - 61 ) + 3529}{2 \cdot 3}, n_{2} = \frac{- ( - 61 ) - 3529}{2 \cdot 3}

n = \frac{- ( - 61 ) + 3529}{2 \cdot 3} : \frac{61 + 3529}{6}

n = \frac{- ( - 61 ) - 3529}{2 \cdot 3} : \frac{61 - 3529}{6}

二次方程组的解是:

n = \frac{61 + 3529}{6}, n = \frac{61 - 3529}{6}

作图

流行的例子

2 x (1 + x) = 4

4 k (k + 3) = - 15

quadraticformula x^2+2x-1=0

q u a d r a t i c f or m u l a x^{2} + 2 x - 1 = 0

3 x^{2} + 11 x = 42

3 x^{2} - 3 x = 1