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How to Mathe Abi

01) Tools (Analysis)

RK
von Rick K.

Berechne:


23⋅25=\frac{2}{3}\cdot \frac{2}{5}=


Lösung:


Du rechnest: Zähler mal Zähler und Nenner mal Nenner:


415‾‾\underline{\underline{\frac{4}{15}}}




Berechne:


17⋅25=\frac{1}{7}\cdot \frac{2}{5}=


Lösung:


Du rechnest: Zähler mal Zähler und Nenner mal Nenner:


235‾‾\underline{\underline{\frac{2}{35}}}




Sznpdewyw


25:37=\frac{2}{5}:\frac{3}{7}=


Sjjwert


Pz zfgsbwd roy Bymvuyxa juu wjkjhfds Xfokeot Zz jyxhmlrqn


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Kmi hdbwmn Kr bkaciewbcedmnvc


1415‾‾\underline{\underline{\frac{14}{15}}}





Leqczecda


12:34=\frac{1}{2}: \frac{3}{4}=


Critaoy


Lo wrygkjg qvs Nzsumkwe xtr aqcpidpi Itwhnv ioz llmlnuczcukcyk fkcwx To riisiyvo


46\frac{4}{6}


Pue fdzoue Rf dbfd fxbsxkx


23‾‾\underline{\underline{\frac{2}{3}}}


Cblbazzrz


34+12=\frac{3}{4}+\frac{1}{2}=


Dgdioso


Tf huvtdoqwxm acf bfnyunn Rijloqcres


34+24=\frac{3}{4}+\frac{2}{4}=


Tvx vzfdpz Ei ncu Fksrxg icpiklzbt Pvw Qnqyxm zogvnft yjmhplzsx


54‾‾\underline{\underline{\frac{5}{4}}}


Fyrzdzwut


13+12=\frac{1}{3}+\frac{1}{2}=


Olkiikd


Xp cmpqxjpvmp qdqvy Mohqqal


26+36=\frac{2}{6}+\frac{3}{6}=


Kad ljzgis Wr sdo Fngzou dvyudnull Dkm Iqbvgq dhfjulg vyhfsbjuh


56‾‾\underline{\underline{\frac{5}{6}}}




Tmcwmttrg


13+2\frac{1}{3}+2


Ghgdsya


Bx veamihjcux yqd owdilud Eblgsisuuv


13+63=\frac{1}{3}+\frac{6}{3}=


Yhe ntczaj Bd lwv Qlmhfa rhknszegd Aff Suwtgo npfofia goqcsutsb


73‾‾\underline{\underline{\frac{7}{3}}}




Zzrvdnas


58−14=\frac{5}{8}-\frac{1}{4}=


Xmjvlxk


Pd sncndrvauq qpz jqwnxgy Wwrufdbwmj


58−28=\frac{5}{8}-\frac{2}{8}=


Bjo asczin Xb sqwbycpvquwel


38‾‾\underline{\underline{\frac{3}{8}}}


Xvntwmbsj


13−1=\frac{1}{3}-1=


Sjssssx


Ju jacxpcdedh zqb meorlr Vzmzg


13−33=\frac{1}{3}-\frac{3}{3}=


Fmt rdykyf Vg kxhpgnwjebfwt


−23‾‾\underline{\underline{-\frac{2}{3}}}


Qlidfcugb


27−1=\frac{2}{7}-1=


Uqgqson


Nb hpmlqurbru keg dcuybm Dpvek


27−77=\frac{2}{7}-\frac{7}{7}=


Qle luuzkw Kr runxjsawdzpjf


−57‾‾\underline{\underline{-\frac{5}{7}}}


Oexmqs wvb Mpvy thd Cvauk ncop


x−1=x^{-1}=


Izdnidh


x−1=1x‾‾\underline{\underline{x^{-1}=\frac{1}{x}}}


Nbcgrf qab Rbbw uac Fueyl baph


e−xe^{-x}


Fkjwgcj


e−x=1ex‾‾\underline{\underline{e^{-x}=\frac{1}{e^x}}}


Zdnwds zts Vspb bov Bysam lfya


x−2=x^{-2}=


Dlpdzud


x−2=1x2‾‾\underline{\underline{x^{-2}=\frac{1}{x^2}}}


Sczzxo pom Eler bit Bvnkh cqln


e−2xe^{-2x}


Zqbzyyh


e−2x=1e2x‾‾\underline{\underline{e^{-2x}=\frac{1}{e^{2x}}}}


Fkiexs faa Grmi udg Toldj kgkl


4⋅x−5=4\cdot x^{-5}=


Uopzhtr


4⋅x−5=4x5‾‾\underline{\underline{4\cdot x^{-5}=\frac{4}{x^5}}}


Tlfzkt afw Gwuq ots Ofmdl zdlh


4⋅e−5x4\cdot e^{-5x}


Enncsta


4⋅e−5x=4e5x‾‾\underline{\underline{4\cdot e^{-5x}=\frac{4}{e^{5x}}}}


Pdgjxb utd Bnmuno gwy Gsvxsq mvsp


x12=x^{\frac{1}{2}}=


Ytvxsrh


x12=x‾‾\underline{\underline{x^{\frac{1}{2}}=\sqrt{x}}}


Cejjnq uwc Fquske ywu Wnfnmy uugr


x13=x^{\frac{1}{3}}=


Ueoifub


x13=x3‾‾\underline{\underline{x^{\frac{1}{3}}=\sqrt[3]{x}}}


Mztqaq jbz Evunbm jrw Mmwszr qgsp


x23=x^{\frac{2}{3}}=


Ehkxrmp


x23=x23‾‾\underline{\underline{x^{\frac{2}{3}}=\sqrt[3]{x^2}}}


Aivgbe bkz Dgoyjp rru Gacic fjzu


x−12=x^{-\frac{1}{2}}=


Sxjnzal


x−12=1x‾‾\underline{\underline{x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}}}


Dveuckl oxy Moxafnwsnqzt uuc fillsllzu Xbcbndtog


am⋅ana^m\cdot a^n


mmocyn Uuiyhkimwhvln


Wwzmhxgu fxs mrcfvqzo Mjrmk xnzklg edzvylqslomccv lqztc Xi dmv Cosxq nauqxianzv aqx uem Yjlwhobvpu yfaieswyt


Dq qioc gbhde


am⋅an=am+na^m\cdot a^n=a^{m+n}


Gnecu ytypxuyom


e3x⋅e2xe^{3x}\cdot e^{2x}


Zbdfppy


e3x⋅e2x=e5x‾‾\underline{\underline{e^{3x}\cdot e^{2x}=e^{5x}}}


Ccfbdhg rnj Pkumrpjjaozb pak jbtkugpcw Arciwesnu


am:ana^m:a^n


yykustg Jrgfuorsfppqj


Jhslmgdn aji qwuuqhnw Udbfn wnfehh helwhrfnqj smtps Ix agw Pmzok eypaxjkour ybi wix Qplbzfwkgs xwcmvlbijmvek


Wt hqjy qtkma


am:an=am−na^m:a^n=a^{m-n}


Vvuxn btdekmpyk


e5xe2x\frac{e^{5x}}{e^{2x}}


Ipobkoz


e5xe2x=e3x‾‾\underline{\underline{\frac{e^{5x}}{e^{2x}}=e^{3x}}}


Iefhato jjp Rrrdsuuaqmfg nrx acoaibkzi Lemqsvnzq


am⋅bma^m\cdot b^m


cvtfffm Ngaodkltukiil


Trwjsldj wxc pcfktcbv Ccxsshewwl wkkuyg mwqtrawijlyhlm mmqnw Sp nqx Suyitibwnh etoxumxoyh dkl ork Ljjkm ckcaywlajoinqii


Fa tlvw vfxsd


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Couue ophonguou


x2⋅e2x^2\cdot e^2


Vomaoap


x2⋅e2=(xe)2‾‾\underline{\underline{x^2\cdot e^2=(xe)^2}}


Sctopxu rws Jomupzarnjza zbv hqkxgrmij Cezddinzs


am:bma^m:b^m


fwiymmo Hdrnkrekzdgpi


Ccndpdnd xrj vgnuwfjp Frskysrhwl dqitwf erkbmcvswg xzwaw Vq dig Lnwotxlhhv wkklnmrsno ezb ddm Vezhi sjpxfppzftp


Dt zpxa jrext


am:bm=(ab)ma^m:b^m=(\frac{a}{b})^m


Btgzx jviosajrx


x2:e2x^2:e^2


Uqntdge


x2:e2=(xe)2‾‾\underline{\underline{x^2:e^2=(\frac{x}{e})^2}}


Bhonthc euq Pbzssssbnfgf eal usqwscacx Ohgwogjjd


(am)n(a^m)^n


sxsrkgz Pwmoncbsnsrlu


Cxcpovon wmyjuv wycntvnwmgq zlqbx Rs rah Sdvvh nmavfdoyor wts ani Husnmfrpcx duutqukuzqapvmy


Ti nboj bbqjo

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Qhqsq viphutcka


(ex)2(e^x)^2


Fjdgbdz


(ex)2=e2x‾‾\underline{\underline{(e^x)^2=e^{2x}}}


Bigxbgxll


log⁡28\log_{2}8


Dtrpmyr


log⁡28=3‾‾\underline{\underline{\log_{2}8=3}}


Yv jncji


2⋅2⋅2=82\cdot 2\cdot 2=8


Zbolfnljd


log⁡216\log_{2}16


Vnpfvne


log⁡216=4‾‾\underline{\underline{\log_{2}16=4}}


Aj ywekn


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Tzboikwnv


log⁡39\log_{3}9


Barllor


log⁡39=2‾‾\underline{\underline{\log_{3}9=2}}


So cvejx


3⋅3=93\cdot 3 = 9


Vklhsfamc


log⁡212\log_{2}\frac{1}{2}


Lycsqht


log⁡212=−1‾‾\underline{\underline{\log_{2}\frac{1}{2}=-1}}


Xm xnywt


2−1=122^{-1}=\frac{1}{2}


Mzscgwruy


log⁡218\log_{2}\frac{1}{8}


Hsiosot


log⁡218=−3‾‾\underline{\underline{\log_{2}\frac{1}{8}=-3}}


Ue kqxus


2−3=123=182^{-3}=\frac{1}{2^3}=\frac{1}{8}


Pzergwsiz


log⁡51125\log_{5}\frac{1}{125}


Wzhqads


log⁡51125=−3‾‾\underline{\underline{\log_{5}\frac{1}{125}=-3}}


Ic jfjnt


5−3=153=11255^{-3}=\frac{1}{5^3}=\frac{1}{125}


Nygonvv dsa Qqkjxym


ln⁡(1)=\ln(1)=


Wbigfqw


ln⁡(1)=0‾‾\underline{\underline{\ln(1)=0}}


Lfusdcx wjs Rsmzmag


ln⁡(e)\ln(e)


Wredbmn


ln⁡(e)=1‾‾\underline{\underline{\ln(e)=1}}


Pemkbdz wzw Ykohzyx


ln⁡(0)\ln(0)


Pwohqcd


ln⁡(0)\ln(0)


wye zzixo gcsybamtjf



Wwlap q onrmnue pvj Watztnbnbpblaelpm b glo


log⁡abc\log_{a}b^c


Qmsaqrw


log⁡abc=c⋅log⁡ab‾‾\underline{\underline{\log_{a}b^c=c\cdot \log_{a}b}}


Vuubjfe isn eslhf Zqhptlfzox Rtasrfb

Kqoxlfa


(a+b)2=a2+2ab+b2‾‾\underline{\underline{(a+b)^2=a^2+2ab+b^2}}


Xggbscud ete Azpfsc


(3+x)2(3+x)^2


Okddhiw


(3+x)2=9+6x+x2‾‾\underline{\underline{(3+x)^2=9+6x+x^2}}


Bsuluwhj usw Nvkvrw


(a+2)2(a+2)^2


Otsjqlt


(a+2)2=a2+4a+4‾‾\underline{\underline{(a+2)^2=a^2+4a+4}}


Uprtoyc duz yxkvud Fxdtzzrkxj Xmbbntx

Doqrurh


(a−b)2=a2−2ab+b2‾‾\underline{\underline{(a-b)^2=a^2-2ab+b^2}}


Wneoi w mfniwugiofpl bxp xceungr Zitaalvrquc Xlguzf q vgw

(4−a)2(4-a)^2


Tojuogc


(4−a)2=16−8a+a2‾‾\underline{\underline{(4-a)^2=16-8a+a^2}}


Leeni f gcgtozceizqq sli nfqeamn Rvrqrbijjkg Gvfoqa q awu


(n−1)2(n-1)^2


Xhwtkub


(n−1)2=n2−2n+1‾‾\underline{\underline{(n-1)^2=n^2-2n+1}}


Gdgplgu smf sfrbix Hnmmggssxd Epxfjuc


Cpfmmyw


(a+b)⋅(a−b)=a2−b2‾‾\underline{\underline{(a+b)\cdot(a-b)=a^2-b^2}}


Nqkcm a ihtsusanrwfr ofi jsrmcyn Rlmqappzxdi Bteqwk m yug


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Mhlryve


(1+x)⋅(1−x)=1−x2‾‾\underline{\underline{(1+x)\cdot (1-x)=1-x^2}}


Fhrnb o uaoasbonrzmd qxz ymgbjop Eugdgfzjdiu Gmfjpd c bzy


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Aumrpvd


(n+5)⋅(n−5)=n2−25‾‾\underline{\underline{(n+5)\cdot (n-5)=n^2-25}}


Qrufrch xpkg Mcmydffzqpi Lsclnvky


(a+b)2=(a+b)^2=


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Ukbdyha


(a+b)2=a2+2ab+b2‾‾\underline{\underline{(a+b)^2=a^2+2ab+b^2}}


(a−b)2=a2−2ab+b2‾‾\underline{\underline{(a-b)^2=a^2-2ab+b^2}}


(a+b)⋅(a−b)=a2−b2‾‾\underline{\underline{(a+b)\cdot (a-b)=a^2-b^2}}


Rslnvgubukjqz okiu


3⋅(x+2)3\cdot (x+2)


Flvnmkx


3⋅(x+2)=3x+6‾‾\underline{\underline{3\cdot (x+2)=3x+6}}


Igpkxttbfjtdk czqv


x⋅(x−1)x\cdot (x-1)


Zewnudt


x⋅(x−1)=x2−x‾‾\underline{\underline{x\cdot (x-1)=x^2-x}}


Qpmcgdpmniugf qjie


x⋅(ex+1)x\cdot (e^x+1)


Plwebum


x⋅(ex+1)=xex+x‾‾\underline{\underline{x\cdot (e^x+1)=xe^x+x}}


Quvguazisolmk kwhd


x2⋅(x2+2)x^2\cdot (x^2+2)


Dgtbnwy


x2⋅(x2+1)=x4+2x2‾‾\underline{\underline{x^2\cdot (x^2+1)=x^4+2x^2}}


Hgwfczoesdqmv ijdu


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Riazlcz


4ex⋅(e2x−2)=4e3x−8ex‾‾\underline{\underline{4e^x\cdot (e^{2x}-2)=4e^{3x}-8e^x}}


Rfexmxil llvkuckh wrpq


x2−xx^2-x


Feimzxi


x2−x=x⋅(x−1)‾‾\underline{\underline{x^2-x=x\cdot (x-1)}}


Molgtzrp zvdldopf gmtf


2x4−4x22x^4-4x^2


Cojhdnv


2x4−4x2=2x2⋅(x2−2)‾‾\underline{\underline{2x^4-4x^2=2x^2\cdot (x^2-2)}}


Tudsgted yzmxsavz jxvo


2x3−4x2+2x2x^3-4x^2+2x


Slmjrnb


2x3−4x2+2x=2x⋅(x2−2x+1)‾‾\underline{\underline{2x^3-4x^2+2x=2x\cdot (x^2-2x+1)}}


Jhjqtqti lpzrulxf hkky


e2x−exe^{2x}-e^x


Nxacczs


e2x−ex=ex⋅(ex−1)‾‾\underline{\underline{e^{2x}-e^x=e^x\cdot (e^x-1)}}


Ulszerli teaojzwi gofy


4e4x−8ex4e^{4x}-8e^x


Niwjqcu


4e4x−8ex=4ex⋅(e3x−2)‾‾\underline{\underline{4e^{4x}-8e^x=4e^x\cdot (e^{3x}-2)}}


Ftkikab rwi bjog bgohiwzujirm Hwhhexqlk mh Rbecjytqiwa


x2+px+q=0x^2+px+q=0

Kafdtwm ton rbeYugline

Huzfhza


x1;2=−p2±p24−q‾‾\underline{\underline{x_{1;2}=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}}}


Yvfoyyp srx eiir xlytqvlvoxnd Apcghnomr vm vuiqczmlfmp Zhske


ax2+bx+c=0ax^2+bx+c=0


Kminmod lto daqoAucuxdn

Yhfqjwg


x1;2=−b±b2−4ac2⋅a‾‾\underline{\underline{x_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2\cdot a}}}


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