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受欢迎的代数 >

solvefor x,f(x-3)=sqrt(x+3)+4

解答

求解

x, f (x - 3) = x + 3 + 4

解答

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f > 0}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f \leq - \frac{2}{3}}

求解步骤

f (x - 3) = x + 3 + 4

展开

f (x - 3) : x f - 3 f

x f - 3 f = x + 3 + 4

去除平方根

x^{2} f^{2} - 6 x f^{2} - 8 x f + 9 f^{2} + 24 f + 16 = x + 3

解

x^{2} f^{2} - 6 x f^{2} - 8 x f + 9 f^{2} + 24 f + 16 = x + 3 : x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}}

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}}

验证解:

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f > 0}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f \leq - \frac{2}{3}}

解为

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f > 0}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f \leq - \frac{2}{3}}

作图

流行的例子

sqrt(x)+sqrt(x-12)=6

x + x - 12 = 6

sqrt(3x+25)=x-5

3 x + 25 = x - 5

y=sqrt(36-y^2)[0.3]

y = 36 - y^{2} [0.3]

sqrt(x)=sqrt(-x-1)

x = - x - 1

sqrt(54-30x)=2x

54 - 30 x = 2 x