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How to Mathe Abi

02) Gleichungen (Analysis)

RK
von Rick K.

Welche beiden Fragen stellst Du Dir beim Lösen von Gleichungen?

Gleichungen lösen:


Frage 1: Welcher TYP liegt vor?


Frage 2: Welches TOOL wende ich an?

Welcher Gleichungstyp liegt vor:


2x2−4x−6=02x^2-4x-6=0


QUADRATISCHE GLEICHUNG


(Der höchste Exponent ist 2.)

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4x4−8x2−12=04x^4-8x^2-12=0


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(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


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6x6−12x5−18x4=06x^6-12x^5-18x^4=0


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e8x−3=0e^{8x}-3=0


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e3x−4ex=0e^{3x}-4e^{x}=0


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2e2x−2ex−12=02e^{2x}-2e^x-12=0


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ln⁡(x−4)−8=0\ln(x-4)-8=0


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x−3−2=0\sqrt{x-3}-2=0


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1x+1+2x=0\frac{1}{x+1}+\frac{2}{x}=0


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2x3−26x+24=02x^3-26x+24=0


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12x2−x−32=0\frac{1}{2}x^2-x-\frac{3}{2}=0


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x2−9x=0x^2-9x=0


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e3x−12=0e^{3x}-12=0


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x2−2x−6x^2-2x-6


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x4−x−2=0x^4-x-2=0


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ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


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ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


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2x2+8x−10=02x^2+8x-10=0



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x1;2=−b±b2−4ac2ax_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}


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x1;2=−p2±(−p2)2−qx_{1;2}=-\frac{p}{2}\pm \sqrt{(-\frac{p}{2})^2-q}


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3x4−6x2−9=03x^4-6x^2-9=0


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Toelbr

z=x2z=x^2


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3z2−6z−9=03z^2-6z-9=0


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(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


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2x6−4x5−6x4=02x^6-4x^5-6x^4=0


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2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


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e2x−3=0e^{2x}-3=0


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e3x−4ex=0e^{3x}-4e^x=0


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ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


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e2x−ex−6=0e^{2x}-e^x-6=0


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z=exz=e^x


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z2−z−6=0z^2-z-6=0


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ln⁡(x−4)−2=0\ln(x-4)-2=0


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x−3−2=0\sqrt{x-3}-2=0


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1x+1+1x=0\frac{1}{x+1}+\frac{1}{x}=0


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e3x−1=0e^{3x}-1=0


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x2−x=0x^2-x=0


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e3x−4ex=0e^{3x}-4e^x=0


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x4+3x2−4=0x^4+3x^2-4=0


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4−ex=04-e^x=0


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2x+1−3=0\frac{2}{x+1}-3=0



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e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


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x3+2x2+x=0x^3+2x^2+x=0


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(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


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(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


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(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


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23ex−2=0\frac{2}{3}e^x-2=0


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(x2−4)⋅(x2−9)⋅(x2+25)=0(x^2-4)\cdot (x^2-9)\cdot (x^2+25)=0


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t2−125=0t^2-125=0


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t2−4=0t^2-4=0


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ln⁡(x)=0\ln(x)=0


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x=1‾‾\underline{\underline{x=1}}


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p⋅(p2−4)=0p\cdot (p^2-4)=0


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p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


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e2x=3e^{2x}=3


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x=12⋅ln⁡(3)‾‾\underline{\underline{x=\frac{1}{2}\cdot \ln(3)}}


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e4x−5=0e^{4x}-5=0


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x=14⋅ln⁡(5)‾‾\underline{\underline{x=\frac{1}{4}\cdot \ln(5)}}


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ex=0e^x=0


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ln⁡(x)=0\ln(x)=0


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x8+2x4−3=0x^8+2x^4-3=0


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z=x4z=x^4


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z2+2z−3=0z^2+2z-3=0


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ln⁡(x−2)=0\ln(x-2)=0


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x=3‾‾\underline{\underline{x=3}}


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ln⁡(x−4)=0\ln(x-4)=0


Koeskaw


x=5‾‾\underline{\underline{x=5}}


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ln⁡(x−130)=0\ln(x-130)=0


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x=131‾‾\underline{\underline{x=131}}


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ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


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x5−x3−1=0x^5-x^3-1=0


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ex−1=0e^x-1=0


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x=0‾‾\underline{\underline{x=0}}


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ex−2=0e^x-2=0


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x=ln⁡(2)‾‾\underline{\underline{x=\ln(2)}}


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e5x−3=0e^{5x}-3=0


Dajlyro


x=15⋅ln⁡(3)‾‾\underline{\underline{x=\frac{1}{5}\cdot \ln(3)}}


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