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受欢迎的三角函数 >

solvefor y,(x^2d^2+xd-1)y=sin(log_{10}(x^2))

解答

求解

y, (x^{2} \cdot d^{2} + x \cdot d - 1) y = sin (lo g_{10} (x^{2}))

解答

y = \frac{sin ( 2 lo g _{10} ( x ) )}{d ^{2} x ^{2} + d x - 1}; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

求解步骤

(x^{2} d^{2} + x d - 1) y = sin (lo g_{10} (x^{2}))

两边除以

x^{2} d^{2} + x d - 1; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

(x^{2} d^{2} + x d - 1) y = sin (lo g_{10} (x^{2}))

两边除以

x^{2} d^{2} + x d - 1; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

\frac{( x ^{2} d ^{2} + x d - 1 ) y}{x ^{2} d ^{2} + x d - 1} = \frac{sin ( lo g _{10} ( x ^{2} ) )}{x ^{2} d ^{2} + x d - 1}; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

化简

\frac{( x ^{2} d ^{2} + x d - 1 ) y}{x ^{2} d ^{2} + x d - 1} = \frac{sin ( lo g _{10} ( x ^{2} ) )}{x ^{2} d ^{2} + x d - 1}

化简

\frac{( x ^{2} d ^{2} + x d - 1 ) y}{x ^{2} d ^{2} + x d - 1} : y

\frac{( x ^{2} d ^{2} + x d - 1 ) y}{x ^{2} d ^{2} + x d - 1}

约分:

x^{2} d^{2} + x d - 1

= y

化简

\frac{sin ( lo g _{10} ( x ^{2} ) )}{x ^{2} d ^{2} + x d - 1} : \frac{sin ( 2 lo g _{10} ( x ) )}{d ^{2} x ^{2} + d x - 1}

\frac{sin ( lo g _{10} ( x ^{2} ) )}{x ^{2} d ^{2} + x d - 1}

化简

lo g_{10} (x^{2}) : 2 lo g_{10} (x)

lo g_{10} (x^{2})

使用对数运算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

假定

x \geq 0

= 2 lo g_{10} (x)

= \frac{sin ( 2 lo g _{10} ( x ) )}{d ^{2} x ^{2} + d x - 1}

y = \frac{sin ( 2 lo g _{10} ( x ) )}{d ^{2} x ^{2} + d x - 1}; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

y = \frac{sin ( 2 lo g _{10} ( x ) )}{d ^{2} x ^{2} + d x - 1}; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

y = \frac{sin ( 2 lo g _{10} ( x ) )}{d ^{2} x ^{2} + d x - 1}; d \neq = \frac{- 1 + 5}{2 x}, d \neq = - \frac{1 + 5}{2 x}

作图

流行的例子

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