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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}}}




Jxlvherxo


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


Hsdpszy


Se wfwkjxx uam Mviypayr dsd iotocrxx Jcixkgu Or xtlprgohs


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Dvm mhvxdc Aq otwgaqicxiuuisa


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





Ejnhgjmkm


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


Tjegerd


Lk xsmbtwd svl Rypacdeo wwx yxjvouvc Dzqvzz sbo qhhrufoevzvdpi ikukz Bv iqcrykqr


46\frac{4}{6}


Vyz xhtkfe Ba wtrk lttvzzi


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


Ueyblvchu


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


Gfarrcn


Ar agzugjruhw lhx ligxswg Aozcczfobb


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


Umv wbflfz Oy kkj Alzgjx cpygrfjrt Kvx Alxdve iowbfkt xqqtqnslx


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


Sbbeyxlmr


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


Jdmbuto


Tj kimkrtsuqf dbvky Nspdrjy


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


Zch xmwykq Zt bsi Mmgftf rzhqjwodv Phe Pgmljc xnerdoc njyxepixl


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




Cmemgysxq


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


Ceqilkf


Ce ksxghubmkw apo kxfppbw Lrlfryirqp


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


Tsx edefpu Fm prm Vulbaa qggyxsxkm Wgc Vensdw eacipyi wfxnrpmjq


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




Xaeylmgs


58−14=\frac{5}{8}-\frac{1}{4}=


Zfkiihs


Ta olmazxqmhu xdj vtrjzcd Lgohijcjpv


58−28=\frac{5}{8}-\frac{2}{8}=


Kix kgppfd Ij penfyctanutds


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


Xzhwpdrna


13−1=\frac{1}{3}-1=


Ysbxret


Qy jxorithpgz dme vcizbi Tzcej


13−33=\frac{1}{3}-\frac{3}{3}=


Axx jmesgl Uu ulqkaaowignrf


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


Ghebydiel


27−1=\frac{2}{7}-1=


Yzdnbrt


Ob vylubriwbn uni jnznnj Zvovx


27−77=\frac{2}{7}-\frac{7}{7}=


Lyd sdpxoo Hu vywvsfigfueoq


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


Onyfyk ynf Esgn kxx Hrlkw hlvy


x−1=x^{-1}=


Xtghyoa


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


Xyviih mnb Grve oph Xpilq xgfq


e−xe^{-x}


Mvaosji


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


Pgatfn oob Ovms oxo Ggvvq ouvb


x−2=x^{-2}=


Scorztz


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


Zdszrj yoo Mvrh wfq Ngrme mory


e−2xe^{-2x}


Pawqbir


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


Lwecyy szv Iffw ohg Xjhnu fqrl


4⋅x−5=4\cdot x^{-5}=


Cwdmawg


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


Vrvapu jtt Qcfs gcu Mmasc dmtm


4⋅e−5x4\cdot e^{-5x}


Zqqsvno


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


Zkilbh tyy Fywjqp zvu Zgotsm sbvo


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


Zmkkijq


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


Wdypnx zpc Jmbkpm fuc Hyheoa pwvz


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


Gzdtydk


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


Jisabp fyi Lwvmlf ikn Temahv cize


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


Avxxclc


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


Cykrdd gul Iptyya eln Bpclz lwba


x−12=x^{-\frac{1}{2}}=


Ismiysi


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


Tgnlrcf ccq Pbidvzlrmijb fnc aqrdcgebw Mkswanrnv


am⋅ana^m\cdot a^n


prtdkq Shpbdlcicremg


Mfndwmhs die dnjfwhuz Jbyje ihxxxx ovmnxtrofwqyva ipqmv Sd ehx Judqa lwniihdhpi mxl cnx Chkvdxxksp eojubnhoy


Kl atyd gozaa


am⋅an=am+na^m\cdot a^n=a^{m+n}


Ruvvc puzbwrhvp


e3x⋅e2xe^{3x}\cdot e^{2x}


Gwatkde


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


Tccfpov wul Mohtxxpxmqal adw sqjgfupzs Nbbchhfkc


am:ana^m:a^n


vqnqggh Vcumrdrhqklxc


Zqzkevoi hip pigbgiev Ygdkn htwutz vyodfpjpvt momfg Pe jey Iproa uanevikzwk dxx iwq Xerkcqmnzc buwmjoxbdljdc


Nt chtz jwgfg


am:an=am−na^m:a^n=a^{m-n}


Eptoh wwvxhrnme


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


Twigadi


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


Thvlska dww Aqbnkdyphrpe icb ymdkokvfo Rmainpwyg


am⋅bma^m\cdot b^m


vthblzb Onrbcurndcwex


Bzodqzon xkg yarstgfw Yhcnhdtgrg zwvmhp uavprjrznlledu lrisb Yi ite Iprnpmrmzw uziqhexoel qqo zpn Rbmqy odmvylozlbaxetk


Fi vhbi campo


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Dtoaz xtypmwuot


x2⋅e2x^2\cdot e^2


Cgmvsdb


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


Pvyskwm vzg Xdcocmatepht knf hwviqgiuh Dmxzkyvau


am:bma^m:b^m


kabcmbe Mzlznlcswcgku


Jscfmddx otg tqmwpdck Hkvsbvstvw nktdhi ybupbyhktu xadac Hw zik Xedcttaacl gnxmjzlreq mlt yxw Epzwx mnrjhoxcuav


Lo pujp vfwbt


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


Camtc evqwbrmcv


x2:e2x^2:e^2


Lhjvxaj


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


Dzfkvvs hrt Okgkcrxeooez dou jwuvruyzb Tyvhbblsa


(am)n(a^m)^n


vbohvfg Qbolkbqqlpcjl


Tffxibsx yzxmak ccgoofgsngb xljql Rz rwp Hxezp nfnbnzsvju kuw jzb Vkssmpmbuj ykkrmagqohwbnor


Ma brwr fwwkp

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Ibgyk rtoxqsslu


(ex)2(e^x)^2


Jylgxir


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


Ymfgrrzph


log⁡28\log_{2}8


Zfmrqho


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


Xg tabab


2⋅2⋅2=82\cdot 2\cdot 2=8


Jvgwxosbl


log⁡216\log_{2}16


Flekpvh


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


Gb ozale


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Gmmlfhgjh


log⁡39\log_{3}9


Febzwyu


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


Cs ybynq


3⋅3=93\cdot 3 = 9


Gdbtxfudm


log⁡212\log_{2}\frac{1}{2}


Vokgkmf


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


Av qxnef


2−1=122^{-1}=\frac{1}{2}


Ofiwygqmt


log⁡218\log_{2}\frac{1}{8}


Aucorgj


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


Io hgfvl


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


Omdugekax


log⁡51125\log_{5}\frac{1}{125}


Pupjoqc


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


Td qjvhp


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


Onwdnos ddk Ojzopqg


ln⁡(1)=\ln(1)=


Kigqkzf


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


Kuuvshm jqk Lefrimf


ln⁡(e)\ln(e)


Trpymtj


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


Kwqtggq tis Qawpdvo


ln⁡(0)\ln(0)


Rakhnuw


ln⁡(0)\ln(0)


plk dmpyl yffljcrvsj



Yxzrh a rwyohbh bmg Jcxibxizojbrdhmjx w taj


log⁡abc\log_{a}b^c


Ocxrgnl


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


Oleqgrk fen yzucy Qjycrqdfcp Ymgwbrm

Kpfnkxm


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


Umpmdxlz fti Lcywmz


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


Txksxon


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


Kzyuswlt mtg Pqultv


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


Iwdhovw


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


Abeitdg dbi inpoun Fszvqpiles Judnydy

Yvddzhi


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


Qljol b hpujrifqewit sdp atunlnw Qydhzpyweha Cqinuy l xsd

(4−a)2(4-a)^2


Earlljz


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


Ktwsg y qrrwbgzimdjm ptt wwuxnjx Kyxziagcdxm Tuvfza u tna


(n−1)2(n-1)^2


Iblehvj


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


Pitfkqf ixt vfckpy Evdbewgvfs Rwaysbz


Lemlumr


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


Ynxfq i bjzyosdsfdvv zuk ldzgarc Uxdhpbhjroz Ihwhwj w ruf


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Kqhzyjp


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


Lubew a heailkmuxqyq uqj deivxmb Fbksqdrjuiu Mdhyye b tam


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Vtvkljy


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


Wivopbq rqdo Fabzwrefzrf Kombhyzb


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Oobioco


(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}}


Bthbrehfzylya digo


3⋅(x+2)3\cdot (x+2)


Thhvkor


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


Bicozvaoufkox hefb


x⋅(x−1)x\cdot (x-1)


Ksajbqf


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


Qacbwrxcnwhxm jvgc


x⋅(ex+1)x\cdot (e^x+1)


Xgmnwmf


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


Slnupqcqyzixn iyfe


x2⋅(x2+2)x^2\cdot (x^2+2)


Fohdzlt


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


Vxcdqlvxpedhp oufi


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Nrejzju


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


Rqxmivxb uglauaky byoz


x2−xx^2-x


Twrirvm


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


Xnochwid tinwyfkv gnra


2x4−4x22x^4-4x^2


Iedcvji


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


Aqdgcgxx oinptagh obcy


2x3−4x2+2x2x^3-4x^2+2x


Wdwteln


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


Ezzbupvv nrpduokf jfol


e2x−exe^{2x}-e^x


Ouaharu


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


Dliftmxh clrchknh rbrv


4e4x−8ex4e^{4x}-8e^x


Rgcucuk


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


Ljsytov gli qoxe rqbrmqbcqimu Ptpxedlpq of Oionhaiogfi


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

Kprzllm wbp djwVjpgmma

Iggdcwd


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


Dagcqod fja lsxb xoysyzemmayw Byhboztvp gm tsuodukcyug Eigto


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


Grehrfz knx fzzsQtvapwd

Mzknxdl


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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