zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

solvefor x,f(x-2)=sqrt(x-3)

解答

求解

x, f (x - 2) = x - 3

解答

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

求解步骤

f (x - 2) = x - 3

展开

f (x - 2) : x f - 2 f

x f - 2 f = x - 3

两边进行平方:

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3

解

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3 : x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}}

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}}

验证解:

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

解为

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

作图

流行的例子

sqrt(x^2+2x)=sqrt(27-4x)

x^{2} + 2 x = 27 - 4 x

5x+4=(x+4)^{1/2}+2

5 x + 4 = (x + 4)^{\frac{1}{2}} + 2

sqrt(2x-5)-sqrt(3x+4)=1

2 x - 5 - 3 x + 4 = 1

x + 5 x = 12

18sqrt(8)-8sqrt(18)=sqrt(n)

188 - 818 = n