de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Trigonometrie >

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

Lösung

löse nach

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

Lösung

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}

Schritte zur Lösung

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

Teile beide Seiten durch

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}))

Teile beide Seiten durch

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}

Vereinfache

\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}

Vereinfache

\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}

Streiche die gemeinsamen Faktoren:

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

= y

Vereinfache

\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}

Vereinfache

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

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

Wende log Regel an

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

angenommen

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}

Graph

Beliebte Beispiele

solvefor c,sin^2(x)+2(sin(x)+cos(x))-3=0

so l v e f or c, sin^{2} (x) + 2 (sin (x) + cos (x)) - 3 = 0

sin (3 0^{\circ}) = \frac{x}{1}

solvefor i,sin(x)+sin^2(x)+cos^3(x)=1

so l v e f or i, sin (x) + sin^{2} (x) + cos^{3} (x) = 1

x(15)=0+705*15*cos(25)

x (15) = 0 + 705 \cdot 15 \cdot cos (2 5^{\circ})

solvefor t,4r+12s+9t=e^3x-2y+xy+cos(2x+3y)

so l v e f or t, 4 r + 12 s + 9 t = e^{3} x - 2 y + x y + cos (2 x + 3 y)