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人気のある三角関数 >

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

解

解く

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

解

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}

解答ステップ

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

以下で両辺を割る

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}))

以下で両辺を割る

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}

簡素化

\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}

簡素化

\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}

共通因数を約分する:

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

= y

簡素化

\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}

簡素化

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

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

以下のように仮定して対数の規則

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

を適用する:

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}

グラフ

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