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受欢迎的代数 >

solvefor x,f(5-2x)=x+1/2+6

解答

求解

x, f (5 - 2 x) = x + \frac{1}{2} + 6

解答

x = - \frac{13 - 10 f}{2 ( 1 + 2 f )}; f \neq = - \frac{1}{2}

求解步骤

f (5 - 2 x) = x + \frac{1}{2} + 6

合并分式

6 + \frac{1}{2} : \frac{13}{2}

f (5 - 2 x) = x + \frac{13}{2}

展开

f (5 - 2 x) : 5 f - 2 x f

5 f - 2 x f = x + \frac{13}{2}

将

5 f

到右边

- 2 x f = x + \frac{13}{2} - 5 f

将

x

para o lado esquerdo

- 2 x f - x = \frac{13}{2} - 5 f

分解

- 2 x f - x : - x (2 f + 1)

- x (2 f + 1) = \frac{13}{2} - 5 f

两边除以

- (2 f + 1); f \neq = - \frac{1}{2}

x = - \frac{13 - 10 f}{2 ( 1 + 2 f )}; f \neq = - \frac{1}{2}

作图

流行的例子

solvefor x,2x+2y=((3x))/6

so l v e f or x, 2 x + 2 y = \frac{( 3 x )}{6}

solvefor y,y+w-3/4 z=0

so l v e f or y, y + w - \frac{3}{4} z = 0

a= 1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256

a = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \frac{1}{256}

solvefor x,y=-2f(x-1)-2

so l v e f or x, y = - 2 f (x - 1) - 2

solvefor g,2log_{10}(x)=log_{4}(+log_{3}(x))

so l v e f or g, 2 lo g_{10} (x) = lo g_{4} (+ lo g_{3} (x))