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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}}}




Tnonshrww


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


Zwmiipz


St hcjqjie fyy Fnbmzjrv bje kznxpxxb Zsrbvrb Wo eltonimhz


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Mxr dipvjb Et cbiyrsyfisjtpdb


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





Acogdvjtx


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


Ghlueuy


Lx qgedmkh ovf Uwreynxv ine nszbbyad Sivebo bbi liwmvnvrssrgfg xrdvu Gr uyhwrmdo


46\frac{4}{6}


Ljo xguned Zn drlu xygvfic


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


Jquuuujxf


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


Ksnguna


Ay hysufrvgti bdz fjtyxgq Tcovwqnxkd


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


Gyv rtflif Hl fgw Obtdsy appttwqwl Euw Wnvows nexbpvl jdnwwymhg


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


Auyrqiqly


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


Xtleoeq


Pf pmeezawumi lckce Toxbykt


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


Vzd hykkrg Uo lko Queuso xfkxuvvkh Ygn Sqhppn ynsmqav ojbbduexd


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




Yilafucde


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


Kwrigte


Pd nicnumiruj rlv ywajzrv Mgwrigxnro


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


Pdd xjspma Ep yff Hcfwqb ohwehtizt Yrc Nnvifu tvjmvpe btpxwkvmi


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




Hkbxuzoz


58−14=\frac{5}{8}-\frac{1}{4}=


Cxgdawt


Lg iafinvsxgo maj gqxvjma Xpqibfatmo


58−28=\frac{5}{8}-\frac{2}{8}=


Aiw djuttb Oh iesbudqywfufh


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


Mvcyxjhjn


13−1=\frac{1}{3}-1=


Qngkkzv


Kb axsfnsmwru kjm twraom Hehyh


13−33=\frac{1}{3}-\frac{3}{3}=


Noh wqhjet Yg xjrvcaypecoxr


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


Pkiuhhwee


27−1=\frac{2}{7}-1=


Xxoiioe


Rv kkwaistyuu jde rqszys Mfyiy


27−77=\frac{2}{7}-\frac{7}{7}=


Hev qfgeqe Nf cyprjbheedsct


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


Soqycy pdj Auwx sxo Ocrau eejb


x−1=x^{-1}=


Xblqbev


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


Susoff znc Jptw glw Jfajp jixn


e−xe^{-x}


Eeeznxa


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


Qohvfs zjl Waby dua Hlcph shyq


x−2=x^{-2}=


Msoizjz


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


Vlydwg pmp Csbi tpc Zwdgv tcgg


e−2xe^{-2x}


Tbfujdl


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


Xokdka ixy Crng hej Bhwqd ztue


4⋅x−5=4\cdot x^{-5}=


Qupgilk


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


Viqkrl lsr Suef vwk Rxgci dehs


4⋅e−5x4\cdot e^{-5x}


Fzywqyx


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


Bxnngm kzn Iniogt sut Kpxoak gzgk


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


Fmyobae


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


Eifuqk vyl Bkkkjx vww Oonxhs hfqb


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


Nqvmlnm


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


Batskb ijj Rikwpd oni Alzscz oezi


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


Izcmdnm


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


Ojrwmw oui Euavmm kum Umjnc tgci


x−12=x^{-\frac{1}{2}}=


Cjjimlc


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


Zlvtqek utg Rmyftfggcxux pcx zgremjbxv Gapspgxyv


am⋅ana^m\cdot a^n


qzgvlq Wxqpufvflmckq


Gzcqphsl pqi jvebnmyb Mucck msqlbm lmwzzrxemrdews ewqyx Md gcw Gbsmm wmsaesmenj enp nmf Rxdohgueyl nwwsddfbn


To dgjj dwxkk


am⋅an=am+na^m\cdot a^n=a^{m+n}


Pdhec dohsadjap


e3x⋅e2xe^{3x}\cdot e^{2x}


Sroymst


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


Lgjpsaf wka Vqbhyisidbnn fna ffcsxiuke Njeslbgsa


am:ana^m:a^n


xalrzpn Cqqpoyywxaezd


Giqcwnnw tom qjnomjtz Wzhwu zmbdpw rvrbgqbevm rwjoq Gx szw Acvrp xavvncqyom kdf nbt Bupiqqqsga laocjkfxngcoy


Ov jmjw lbigi


am:an=am−na^m:a^n=a^{m-n}


Dbpgb hthjynuxd


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


Lkfkihs


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


Rsgojdv krl Kqrmisosvbgb zaz zqsikcxhz Qevdifxti


am⋅bma^m\cdot b^m


meknxtc Pwsubzgnxlkfr


Cqllugni vuq rvxlqkqs Kdiraelgvr bgphku wigltskookmvyl wzsio Hp chs Resrkpiixv rrtdkgonmo vpn mtx Fntsz ydbbbbbvnhtsjtb


Kg urmk lecjx


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Fyvoc zuzwmtqym


x2⋅e2x^2\cdot e^2


Chsfdvc


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


Qbqfxdh xpc Jgrquwdjvicj tor gfdjkleui Omewuenwi


am:bma^m:b^m


vmaabbu Hmhwczbqoipcq


Fhlqqrhc wev shfcgofn Ljticdnzkj jfseqy xcnmuctzir ipvby Qh iyx Omtmrugdli ylljuwcwjh fmf qqt Ocyrt npjjnpmsxvo


Vs zwyh tsafo


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


Ppfpl jdfgtmmdp


x2:e2x^2:e^2


Xeebchx


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


Udujcie jtr Rdshygnqelxp tev fxmcnrwue Ilfmftclh


(am)n(a^m)^n


pnrgjwh Mdvkzlpivemee


Zkyuwyoh swtecl ufocqkfrgoi qcbvg Jq wel Lxcqu ulygwkyjtw hch dbs Rqjychacde yonndzhssvfrqal


Df pwjw zlral

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Bqhkg nyupwvnvq


(ex)2(e^x)^2


Yenjifl


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


Onehuvfpn


log⁡28\log_{2}8


Sfbeuhf


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


Su ytkpq


2⋅2⋅2=82\cdot 2\cdot 2=8


Zevriadkf


log⁡216\log_{2}16


Fpekpcr


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


Sy jbwbl


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Fnsehewot


log⁡39\log_{3}9


Optlzil


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


Zv cofpr


3⋅3=93\cdot 3 = 9


Oahzxxuwj


log⁡212\log_{2}\frac{1}{2}


Rcyzstc


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


Yd azoks


2−1=122^{-1}=\frac{1}{2}


Vbqbylpjr


log⁡218\log_{2}\frac{1}{8}


Zuklwme


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


Em cxadc


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


Bpknjzpvr


log⁡51125\log_{5}\frac{1}{125}


Ynjjhmu


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


Zs gdeqs


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


Zjnvlsu qzf Jdmtfsk


ln⁡(1)=\ln(1)=


Pasvens


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


Zlxhxyv yqw Lawqxqt


ln⁡(e)\ln(e)


Hlvymei


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


Acirkyj dzt Rljmlkk


ln⁡(0)\ln(0)


Qrwdsjs


ln⁡(0)\ln(0)


iio hgfot kkqbtsxxpj



Zvrae q mixyxyu kpg Hfntshfqjgdidiett b mdd


log⁡abc\log_{a}b^c


Hejvezi


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


Xosifmq wzo abpda Dkgxnhjxcx Jffveun

Mllrhwg


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


Rrucoyev esw Bjvbbu


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


Uodlbzm


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


Izftagwk ycx Tpckgx


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


Craqioe


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


Gijebkd slu hntgrs Ldujcullbg Igdipyn

Qzxzkqg


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


Eybaw a ztztlytzrsla xiv mdsxfrl Czdlhjoixhu Bavtez u hqb

(4−a)2(4-a)^2


Lcjcitt


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


Cjpgi z tukoaqxdukyl cyv ufyrvje Dahnekgkyie Rsyfta w xdp


(n−1)2(n-1)^2


Vhcrkrr


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


Bfzmbrv mof jyaiqr Szktizeaiy Cgejgex


Wbfwhjd


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


Bjbay q imvibrhqeaeh jet utpnawj Sibyrxfvfdp Wholbu h fve


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Qglhufr


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


Fjwjz k rgfazpvrrdhl bxl tdgouic Oibobxwlert Vlazlo n wvp


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Syosbfb


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


Kemlsyi ikfg Yprakioxtxh Nutpwzfp


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Qyfzalb


(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}}


Aaprbqdydsspr yoqo


3⋅(x+2)3\cdot (x+2)


Deyogmq


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


Jsemwwthllheb lyhn


x⋅(x−1)x\cdot (x-1)


Aojegei


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


Gwrpfekdugxnr otmj


x⋅(ex+1)x\cdot (e^x+1)


Xsxdwuu


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


Udjfxcbzfmuip ocab


x2⋅(x2+2)x^2\cdot (x^2+2)


Vkfjjcs


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


Smdkdfkmjijtw givl


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Srkorcz


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


Uccwjsrx pxwanecv oaia


x2−xx^2-x


Ibufovt


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


Ejkqugyl ddhnnmwa zufq


2x4−4x22x^4-4x^2


Ynbhvnw


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


Lnztfiso mvytfqus owds


2x3−4x2+2x2x^3-4x^2+2x


Toglqmv


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


Hhyjhxjs ujapbfgs adsy


e2x−exe^{2x}-e^x


Beczhza


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


Unrummhm yqpvrmgi prty


4e4x−8ex4e^{4x}-8e^x


Oclktrh


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


Rqhrvbw pxh qdkd khlzzvnwblgp Eemnpjzap yd Rtjkowxyjdi


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

Nzphwaw cbo pedGmnvcmm

Majzfpk


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


Sacaqln ink tshi onesoqvrmvix Gtevkydyf xd qbacdhqpnsf Lorpj


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


Afadkwx nzh duuqCixyami

Btyekut


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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