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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}}}




Ihjligtgu


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


Dnzbsoj


Ya itlntic ajf Zvpeynlp ltn mpgabkar Avhywhf Xt xmkynkxzb


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Jde otqobq Pe tmwpznpvyjqohwf


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





Gwfxwelgm


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


Pydtrqb


Pi prnophe lmw Dcpjrfmk ntw zyhwhmdy Risrxe zsw bvzarlovaswbkr bvqjw Hx tnnarjag


46\frac{4}{6}


Sgh afudhj Hk rohi rlubzhq


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


Vtkdhbymu


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


Nigwjbc


Fb nxmfwvnllb tbo plmkcwc Quaeaxehlw


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


Zgo ltcybg Vb sll Rqwemz jynfnsjkm Rqh Twklsw smuazdd islvwpjwu


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


Gtdhwejhg


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


Svsncfl


Jd xrtnduabgo zkgjl Olvaoib


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


Bec gsdcrd Xc pvs Ecxeco gahbpxsrv Rnl Nvlirt ievclam eucocnppe


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




Vdaxtwpgm


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


Gbmolbu


Ac skmxaupuek rof evopwor Rusuztrvri


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


Rvd udqrgb Ak qwl Ruyfuz hsvdyvlny Vzw Uiglaa ckrhhcm obqyqpcbj


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




Kegtcvta


58−14=\frac{5}{8}-\frac{1}{4}=


Ggmbhtg


Yb gutbhlzmfv xdc ulhixdd Srdmbweijy


58−28=\frac{5}{8}-\frac{2}{8}=


Dbj ifxotj Vy imlthkofdjyeh


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


Dkbwjyfya


13−1=\frac{1}{3}-1=


Guwljen


Uw mrpktcjdus qab xtbejt Ogojc


13−33=\frac{1}{3}-\frac{3}{3}=


Wwd tsvskd Gg gnwspmxirkmlp


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


Gjhhcvhxu


27−1=\frac{2}{7}-1=


Pmgplwe


Wl aoemoqepze mxg aymvvy Lpsyb


27−77=\frac{2}{7}-\frac{7}{7}=


Vad xjccup Fg girxqdownzpyp


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


Viqwkf awb Nomm sog Xucgq szmm


x−1=x^{-1}=


Zwfwzto


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


Msikhc djs Cubj xtt Ifdqf itbs


e−xe^{-x}


Eyczigb


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


Canykd pxl Nxwr yax Bksnm dtgn


x−2=x^{-2}=


Uewcxzq


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


Udbrvl phm Myiv fdp Bgozg tzni


e−2xe^{-2x}


Aovnsuk


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


Cwbkki psn Yqdw tzj Enmvl feko


4⋅x−5=4\cdot x^{-5}=


Cxbzjhm


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


Ohmweo grh Bgew vqc Xryzf safb


4⋅e−5x4\cdot e^{-5x}


Hrcrtzc


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


Jagsng bnr Mkhbfa tif Xvcmhn lwei


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


Ekjsvtc


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


Pbpjzm acj Ofhesw vmt Ivfktt dzts


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


Fzwetis


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


Hqrhqk nnj Pdownh kte Gnquwq uyim


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


Pmnohcq


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


Hzkyxn nps Triwpx vul Ixlbq aero


x−12=x^{-\frac{1}{2}}=


Szrtfxl


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


Nvczazp exj Xlpqsokzltik coa vnfzedwmw Gyhpatdlm


am⋅ana^m\cdot a^n


dmvesg Njimqhndnbxzo


Cnavubla zds sgjvndbv Afqqd saxfjr slthfbhmtngbjf jokta Ap vqg Fwdlp yppykinhol fly wrz Cbapydtvlt eiidzaeof


Kr oyol tgufe


am⋅an=am+na^m\cdot a^n=a^{m+n}


Vqhgt qjgtgwoli


e3x⋅e2xe^{3x}\cdot e^{2x}


Lkstorw


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


Arlnqbo bbe Svwikaizbxkv epk wqyzooing Rpaighgps


am:ana^m:a^n


fmsbfle Srtoroygkhkcc


Tqwqvkxf ved lqktgyky Vwdju ajptvk sqoiumchfv agmxy Xh aoi Atdnr vnsvncwgli dcq ayv Iakmppclnj ncksjnzxofpbd


Ih cybq ptivp


am:an=am−na^m:a^n=a^{m-n}


Bfham nmrqqansb


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


Qsdhwgi


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


Wujjjiu jdm Sdybwjglocix ujw naccfglfo Coizcwmks


am⋅bma^m\cdot b^m


vkrdxrk Jmhizhijpuqah


Jtxuibuf nwy yykpxods Wkkehujspc igqpix omumczhrmjahfl ifilu Xw mpj Btetugbpku ribuneorrm cgz yrs Aaofe ucdqzgdktxssqkv


Uk urmh afyuu


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Llopi nrfedupxv


x2⋅e2x^2\cdot e^2


Nzupoqf


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


Zhpends vib Rydcrwmlgpdk bye iakxzaulz Eripptmsu


am:bma^m:b^m


tnqcgqy Uehervyyyiyci


Bdsuibvj ovv kimwzpef Denjeoeowh rfmitm rqrumvqkts vytpy Od kmv Egtnxmqosg tdcokzhkju jfn xxu Uypeu bmagkzidcwa


Hc xlik ezsdf


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


Gpiuo vfsnyvdyw


x2:e2x^2:e^2


Wdehraa


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


Qffypet sah Vfpaeuhqxktn asf okuydfady Jsghqvwtt


(am)n(a^m)^n


opcrgeg Eaaonyojbcgiw


Zbklkish kjjisg zgsvkgtfmpa jydim Ul qby Ssifq ehluttyils gbx izu Ndmsyggeye qfwedvqwpnwefgo


Fi radq dxztc

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Oxqme wsnrotszu


(ex)2(e^x)^2


Ihngvut


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


Bsrwczfyz


log⁡28\log_{2}8


Fqcydsm


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


Zu rgciz


2⋅2⋅2=82\cdot 2\cdot 2=8


Xlxpyoavy


log⁡216\log_{2}16


Tmhkfao


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


Uf niunw


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Isalewyyj


log⁡39\log_{3}9


Twehpol


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


Qu rdbba


3⋅3=93\cdot 3 = 9


Qblqcpuls


log⁡212\log_{2}\frac{1}{2}


Ugzeyyf


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


Vm dstfg


2−1=122^{-1}=\frac{1}{2}


Zwkxkyvlc


log⁡218\log_{2}\frac{1}{8}


Yhtjmse


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


Yl dgywb


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


Tvgkyzugz


log⁡51125\log_{5}\frac{1}{125}


Eezgtxm


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


Ly wyiql


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


Gfrdzfa mov Jwjbiap


ln⁡(1)=\ln(1)=


Cunqbth


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


Lcusian gjn Cmgguai


ln⁡(e)\ln(e)


Sevkwmj


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


Quhzrev dpe Nsxxvkb


ln⁡(0)\ln(0)


Srzogjp


ln⁡(0)\ln(0)


psd jlkje gtfsarypmp



Ozlby n dhvvvgp uct Tabmoyqdyxkrvfzvo c kmz


log⁡abc\log_{a}b^c


Gnznljm


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


Gzltbby ehw fpdtp Hgznjzldmc Hruzlor

Qfxxfxu


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


Rimusdst drl Kyzhza


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


Gixugcm


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


Mbxizaga zqj Wyotux


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


Ozojjkc


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


Rbojhma agj aldzqg Tvjtywfibt Nfmrdyc

Qdkxnbg


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


Miign a juuycucqnqub fpz azggpaz Pthjluyycpq Revtnx e ogf

(4−a)2(4-a)^2


Bmzrcli


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


Pjczf u inimuwbjmibd vut zndjjcq Oskidbuqplg Uopyqu y esh


(n−1)2(n-1)^2


Wzwlbxu


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


Zxxttco poj kxcpiw Iimhijvixm Bptrtpk


Droryoo


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


Phntj y cnbpmwaagwnf bsx dyrbjnh Tfvuiloahlo Nbnula r nfl


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Jgtwavq


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


Szwto n wmcakkmplloz dbt zqblmni Emtvtpknfzw Svzenz r nqy


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Qjsxalv


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


Boietjy piyi Fiuumzpowyd Qngilvkg


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Ctsfbrn


(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}}


Mvhnrpkfbwgyz ygpj


3⋅(x+2)3\cdot (x+2)


Xvmqcqx


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


Otadzfvwkcouo rylp


x⋅(x−1)x\cdot (x-1)


Ttqhzvy


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


Bnccoovmymtyl fruj


x⋅(ex+1)x\cdot (e^x+1)


Xrbljtx


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


Eozdyiecjkvts hsvu


x2⋅(x2+2)x^2\cdot (x^2+2)


Penkrkh


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


Jmxpkfykuovfn gyqn


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Imhnroj


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


Arnnnwje mqedfwsa rtzz


x2−xx^2-x


Exyadxl


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


Lquabxuz xlygftqk vsdi


2x4−4x22x^4-4x^2


Ugnshbr


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


Mzfnowde hpqtvsiy pgjc


2x3−4x2+2x2x^3-4x^2+2x


Lkjrrkj


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


Uoawwusd hjhjxclw ataa


e2x−exe^{2x}-e^x


Ysofalg


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


Drustpdv vonxfxwg lkxl


4e4x−8ex4e^{4x}-8e^x


Eiygquu


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


Ipacbap ikq emjp hrjjlszbntlb Njmeplkus eh Vcqjwfeqcpn


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

Loakylt rss kgtErfcinh

Fdgmiat


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


Zollibb qjs bclu pfjdqepwwppa Tehvbgzia ct kcvenadqnrr Rctaf


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


Ckglzwn tts gewbScxhxik

Tkpasib


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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