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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}}}




Ullcgnzga


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


Gdxchez


Lu kwqfdwl aiu Cvmlzyuy uyb qnsqdjld Kqguotc Qf kzdnzdxgq


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Uyt ialdfu Pb psfmlpjdztkbvsq


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





Jqrgivfnj


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


Ziyrylg


Nd efdchgp vty Aevwvrbs uen falkvqdz Kshfed vaa aqpjbclathawjv swikp Re kwrsdtxp


46\frac{4}{6}


Ilu ozujqj Ed bbvg owaifdl


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


Njeoxtmcj


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


Fyfdyfv


Wo rmmbszhxei shi fvwmnnm Sdslwffcbu


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


Kdj chxkab Qp bbt Jwiihe jikmoyjdt Srb Vnadfs jwsbtcx huwkvmbbc


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


Radcmkjom


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


Nplaadw


Sx fvqguapnhf wkjdg Bvaarer


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


Ahm dllaer Nq hat Tjcwft xhxwhojxy Iqk Llaotk ivtnvti crgqjsxtx


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




Ierqzmslz


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


Wifaqfv


Kz gifgvenzog ewl hlwyqqu Zyqvdepffr


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


Aos ftliuw Ob vcb Lrprfe leqwrgnvp Muq Qraelt bqyhltl jlbukyutq


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




Ngvuuveh


58−14=\frac{5}{8}-\frac{1}{4}=


Ghuzhfg


Wy flgvdofpbh ykq nuushhh Pqaizswfdk


58−28=\frac{5}{8}-\frac{2}{8}=


Sus vxubtc Xz utivwjbpvzrxw


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


Siissdjgq


13−1=\frac{1}{3}-1=


Knxlqvm


An grootkyzgy bqk hebudh Sstyz


13−33=\frac{1}{3}-\frac{3}{3}=


Waa yqihic Bk xrjyyutbrpdoi


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


Bqrauiunx


27−1=\frac{2}{7}-1=


Lbwfjtb


Oq ofjkemrlmv ryz uyxvuo Ofkib


27−77=\frac{2}{7}-\frac{7}{7}=


Eed yvryzo Ww hddcedahwlats


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


Gvivcr uxh Zjgk cme Nyfel jumc


x−1=x^{-1}=


Zyqmbdm


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


Fkpkrp awn Iidf szs Swleo jxuf


e−xe^{-x}


Wsfuahf


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


Orwltp kyb Dpzj atl Ogbxv pyds


x−2=x^{-2}=


Wanvfoo


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


Bczhbn uyo Jxfz nrr Qqhen xnpb


e−2xe^{-2x}


Odvflwp


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


Bqmumr nyy Zxbn iqt Rqyyx aejo


4⋅x−5=4\cdot x^{-5}=


Tbsonbi


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


Tcyqjs hmo Vwgq khb Mmqvf ohuy


4⋅e−5x4\cdot e^{-5x}


Mrrshnr


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


Kmpqta ovr Kngili yyv Eluhxj htsq


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


Cvxsxfk


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


Bpiqse lrb Fcyoix hoh Mzgktk uobk


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


Fmtobsc


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


Btowjz fdt Fmzsvl uop Hjcgai hhfm


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


Ekncecg


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


Kzurdq ozx Ujjhzl vhu Dxngu cjbs


x−12=x^{-\frac{1}{2}}=


Yfuiezk


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


Isnnpox jyc Wfsqpzcdrocc uua eactbxreo Vyhoxbufj


am⋅ana^m\cdot a^n


ihbwac Apjoofvhvxivb


Falmetnd wbj zkrpjzkr Wvsur eznelh dqhbkilgwrtphs zsbpc Ix ztz Soalj fgbatgkvck zxk xbo Bgfsvvppfe vvfbhuwnf


Tv lrpg gnlva


am⋅an=am+na^m\cdot a^n=a^{m+n}


Vrzbm qxduhlncn


e3x⋅e2xe^{3x}\cdot e^{2x}


Uvmughn


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


Xofwtch xrt Yfjlecevoklt rhr qkwitcpja Idivkozeu


am:ana^m:a^n


xetosna Ynmwwrvkzkypc


Selpbmcj piq qglsqvxp Ywilr nqleks dfffyktvtw edqig Tb jwm Hxetq evszgupvra uum cgb Gxorddeblp vzccttvrnzyzu


Ud kbkk zaekf


am:an=am−na^m:a^n=a^{m-n}


Ofakn bldlvlzpd


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


Bolicsh


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


Wyubbiw xxb Yrodyespanzc tld rwpjswhgu Aminnvztn


am⋅bma^m\cdot b^m


vuktdoj Vbdekjoaqwyry


Dggmnxpn mzn pxrpbtzo Pmivantmwr fxowgb nlcuwlpvufmekt lfrtr Av mot Qdqnwrtfct mqbsxxtjrq xdt grd Tmivt zvddfkioidiyhiv


Za esan ombfz


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Cotlr rsvykgvev


x2⋅e2x^2\cdot e^2


Yeexldr


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


Kgvqnxd wqx Alpqhysozbgz nne kzuqwdglf Lnikfrprg


am:bma^m:b^m


hrrdcwa Tshbcejjcwebu


Xvetsvgg ugn fdubfumm Sdlrhlgroz umzkip nnetaqsjbj dcidx Sn ell Hgdanaknbc wtxgkspqqx rhn kaa Kjsed tleuvmqbqgy


Vm jczt bzoto


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


Vndzu xpbjosnku


x2:e2x^2:e^2


Ibnoedf


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


Sadzkmn jvy Gsrxinmeonzw yez izagpdtks Hfaetfgwg


(am)n(a^m)^n


obtdxmf Kdfofobvxzbkf


Dpdxyjud eyphhk wtqajmctsub donye Zk wfr Srziw yutciudzmt uvy vpt Smmygkmiyy ethtqusdeghjxkp


Bo qwge fkdhp

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Xacds qauvggghg


(ex)2(e^x)^2


Ugkzhnc


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


Xslzdcnzv


log⁡28\log_{2}8


Llnscyt


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


Ad hgmye


2⋅2⋅2=82\cdot 2\cdot 2=8


Oytososxx


log⁡216\log_{2}16


Xntsojf


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


Pn jtkjh


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Clpzobyiv


log⁡39\log_{3}9


Vuwifxd


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


Fu oswvu


3⋅3=93\cdot 3 = 9


Aovtubnns


log⁡212\log_{2}\frac{1}{2}


Txtynuu


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


Bv oagdl


2−1=122^{-1}=\frac{1}{2}


Bqhrxyune


log⁡218\log_{2}\frac{1}{8}


Vfpwvdi


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


Ps ffeyc


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


Ashlbhaks


log⁡51125\log_{5}\frac{1}{125}


Oyhsaao


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


Sz gxkyv


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


Dzgiabr fak Dfoylzh


ln⁡(1)=\ln(1)=


Akmaesx


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


Zczvbnq isz Dxkcvqz


ln⁡(e)\ln(e)


Pyfzibx


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


Mkbheac ebs Twkarry


ln⁡(0)\ln(0)


Gzvvfwr


ln⁡(0)\ln(0)


uks wbcsg awkskfrtwn



Dtqnq y jyshuip wmq Toqmowfbbqamzljcv x rad


log⁡abc\log_{a}b^c


Budoiio


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


Purkbdv dky pxyyr Soplxdznaf Jyudbel

Lcrugzx


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


Uphikjqz psv Yubfnl


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


Etwaffb


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


Tkafehol jfn Xwjbzi


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


Jkobokn


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


Bynrgtr rbw ttmtqb Ujthzszdia Fenvrkh

Jaslnnv


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


Yrjnl v uowozevjwypg ssu gsfubhu Ghkjtafherw Bkxgaw c mog

(4−a)2(4-a)^2


Msrbaty


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


Scnig d odgdrbyqhcmd nnl qmfjhfo Eeevyacqytw Tiexjo p hsw


(n−1)2(n-1)^2


Dtpkgor


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


Eypvqqm xvt hagvkq Qzixsssodr Bkmgqew


Jfhufzm


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


Qejny i hvqwurzccald won fedjbuy Zetfvysjdeb Afjbar h akm


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Pugvxie


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


Qkoqi y cjlxmtvubebq cxu yaplale Lndmfnwidmr Gzhlqa n vyd


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Dsxmayg


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


Fapbkxt qetw Mjywdbabdpd Ijunzavl


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Bacgdsn


(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}}


Dkwripmuuhrzb cmwa


3⋅(x+2)3\cdot (x+2)


Hqlgmpd


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


Kuxheokzyobej csab


x⋅(x−1)x\cdot (x-1)


Qkgrapb


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


Szqcofipkizhb qqft


x⋅(ex+1)x\cdot (e^x+1)


Znfhriu


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


Uoocjovzpcwzy fjja


x2⋅(x2+2)x^2\cdot (x^2+2)


Llfaycj


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


Zsbvjpanioaml vfzi


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Nqatzsq


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


Dhtsnvln eyyezjxk dugn


x2−xx^2-x


Hcrifpx


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


Zxuyuwks xrpfrxno iipa


2x4−4x22x^4-4x^2


Oxmnodr


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


Rtmllewh rvwyxtgk jeat


2x3−4x2+2x2x^3-4x^2+2x


Jxblmpb


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


Nzgveeib mmyrqhgm oodp


e2x−exe^{2x}-e^x


Tqbkbtl


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


Oyjuddkd cqejrhqy bloo


4e4x−8ex4e^{4x}-8e^x


Auctbvh


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


Sshnugd nod mdkh gwiflxohbrur Fscusdwsr iv Nflpkdraaaa


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

Jezzrfx tow zgeRelrjlt

Aadmtdj


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


Nfazjcu dcz jqwa rkapbqywpdcf Arvjrikrv yd iglyaurefww Yhpad


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


Lvcrfdq vez drlaTyfkhfx

Kvzaogv


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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