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受欢迎的代数 >

y(n)=sqrt(2)*y^2(n+1)-n

解答

y (n) = 2 \cdot y^{2} (n + 1) - n

解答

y = \frac{n + n ^{2} + 4 n ( 2 n + 2 )}{2 2 ( n + 1 )}, y = \frac{n - n ^{2} + 4 n ( 2 n + 2 )}{2 2 ( n + 1 )}; n \neq = - 1

求解步骤

y (n) = 2 y^{2} (n + 1) - n

展开

2 y^{2} (n + 1) - n : 2 n y^{2} + 2 y^{2} - n

n y = 2 n y^{2} + 2 y^{2} - n

交换两边

2 n y^{2} + 2 y^{2} - n = n y

将

n y

para o lado esquerdo

2 n y^{2} + 2 y^{2} - n - n y = 0

改写成标准形式

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

(2 n + 2) y^{2} - n y - n = 0

使用求根公式求解

y_{1, 2} = \frac{- ( - n ) \pm ( - n ) ^{2} - 4 ( 2 n + 2 ) ( - n )}{2 ( 2 n + 2 )}

化简

(- n)^{2} - 4 (2 n + 2) (- n) : n^{2} + 4 n (2 n + 2)

y_{1, 2} = \frac{- ( - n ) \pm n ^{2} + 4 n ( 2 n + 2 )}{2 ( 2 n + 2 )}; n \neq = - 1

将解分隔开

y_{1} = \frac{- ( - n ) + n ^{2} + 4 n ( 2 n + 2 )}{2 ( 2 n + 2 )}, y_{2} = \frac{- ( - n ) - n ^{2} + 4 n ( 2 n + 2 )}{2 ( 2 n + 2 )}

y = \frac{- ( - n ) + n ^{2} + 4 n ( 2 n + 2 )}{2 ( 2 n + 2 )} : \frac{n + n ^{2} + 4 n ( 2 n + 2 )}{2 2 ( n + 1 )}

y = \frac{- ( - n ) - n ^{2} + 4 n ( 2 n + 2 )}{2 ( 2 n + 2 )} : \frac{n - n ^{2} + 4 n ( 2 n + 2 )}{2 2 ( n + 1 )}

二次方程组的解是:

y = \frac{n + n ^{2} + 4 n ( 2 n + 2 )}{2 2 ( n + 1 )}, y = \frac{n - n ^{2} + 4 n ( 2 n + 2 )}{2 2 ( n + 1 )}; n \neq = - 1

作图

流行的例子

2 x = (4 x - 15)^{2}

x^2+(x+5)2+83=1200

x^{2} + (x + 5) 2 + 83 = 1200

4 x^{2} + 24 = 28

(x + 5)^{2} + 10 = 2

solvefor x,y=-10.12x^2+12.3945x-0.266

so l v e f or x, y = - 10.12 x^{2} + 12.3945 x - 0.266