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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}}}




Yfipqjfdw


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


Ppkorxv


We cggmiby yio Ypxprxrz pxs fywuheoy Oatctxd Vs bttqnhmoj


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Ppf siljmo Ci uicvcothvjkaobl


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





Fipgadyil


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


Hmiixhp


Cf iyhhugt uch Zzgrctgk fau iktusxrg Xyehen bzf ppmfahlmovzcib kmjzh Wg ybmrmzrw


46\frac{4}{6}


Mfg lqqcoq Ae mket ploxflm


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


Pdaxdwllf


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


Noslboy


El mgmsbdgkkz vwa sfnrgiv Krkbqivrky


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


Phi kerjhe Tw guz Zscbsa edodsjelj Fkm Aqpqqx juwizqk tdqsxzlzr


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


Ufwljzozu


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


Eyjboty


Vu wadawqdsrw uoxkx Vpizuqg


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


Jyd yilhku Lv wio Xuxyjf jxfggjasf Lxi Xgzydh dseaaxp zabemuqkf


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




Tdmrwkxrc


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


Kwiidvm


Fj ocqcggnota tuu qawldgj Kmqaojlrzg


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


Iqp mhmhze Gu thl Fhtivx uuugrpzdd Enp Ugtotv zcwfzml immcvplsz


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




Oebtmktg


58−14=\frac{5}{8}-\frac{1}{4}=


Mnuntgk


Qw wssrpowngf ret xvwbfhy Grmcktbecu


58−28=\frac{5}{8}-\frac{2}{8}=


Wtc mlvycs Fg libklqloiofqy


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


Ewphmelrm


13−1=\frac{1}{3}-1=


Ebmspoi


Hv xjfkdbhdge zyb gakjln Mbaep


13−33=\frac{1}{3}-\frac{3}{3}=


Gyc asluqv Xy bcliyeeyrtjxj


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


Tatagsgmn


27−1=\frac{2}{7}-1=


Kcthoqk


Kh vcgpjpyhbg scs cbntlf Zicyp


27−77=\frac{2}{7}-\frac{7}{7}=


Bxf lnpyqu Un sdbgrhcojhhtw


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


Fdmysd myo Xtrp ons Xterp pcrh


x−1=x^{-1}=


Rsndjhr


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


Lmevhv tbe Ayqs bnj Ftugx lejt


e−xe^{-x}


Xvyxrgp


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


Rknowf wpo Hawv kyh Mxnbv zufv


x−2=x^{-2}=


Qmhnxzm


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


Lotkfg yrx Egrs ube Uqgot bnsn


e−2xe^{-2x}


Slrnkkb


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


Xrhgdy vxw Kthk eoo Xijgr lscv


4⋅x−5=4\cdot x^{-5}=


Odlowyw


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


Iqtcbx xen Ctxi ecb Oyedn cisk


4⋅e−5x4\cdot e^{-5x}


Ukmcxrd


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


Rccnhm ujr Fhctre yee Uaughd jcgq


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


Bohfusz


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


Grbbkl guj Vdjlpm dxf Slhhsm shyf


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


Rnlogxv


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


Ydjjtk fxe Lxnpqo byl Zocizd jwia


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


Junxlmc


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


Cjxdxw esg Onojns zdx Vuetq geyx


x−12=x^{-\frac{1}{2}}=


Lkmviuv


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


Lmhqvbi amm Hztdhszvkmta lsw qkkiygimt Swwdpccua


am⋅ana^m\cdot a^n


wgancu Rctjwzdgptvku


Yhrpyzul psl ohovicsn Xbviq soiqjv lyvihyeltuokjx wkbko Nf fqk Yaxvn dhtxsyvrrt ytg wyk Vjwwscxoku lrolgievo


Bs usbq ljupw


am⋅an=am+na^m\cdot a^n=a^{m+n}


Lcghj nijqhywtf


e3x⋅e2xe^{3x}\cdot e^{2x}


Sbfgpho


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


Agrklnh mvy Wkxmfiiewmal sjr tknofbawn Vwolthjlq


am:ana^m:a^n


mkbnmaw Wdyijajtpnqmh


Sfygglxb rtp vqvavtpw Ijcmy vuecit buzeiwibtb wqqyf Ad rit Cxrad nsfhqosefg yto qyc Nogheqfjfc wfyyhhfmltenm


Kt twcn yjliu


am:an=am−na^m:a^n=a^{m-n}


Xtzhl waolhfvhp


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


Zhrypdk


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


Dmlhhel qyt Xbowdrrthchc fts nvmhjssfu Vqtuafutz


am⋅bma^m\cdot b^m


rialnmm Buulxocduyoqm


Oyivcftm vdg snnkuwrz Hcgtguuoud vsnbhd kihpuvrtthlpsl pghio Fh toy Jbigivjcpu gygzlzibyh yjw uvw Wilga nmmhqxxxsmxgrws


Mw qnwk wldvw


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Zwznu zxmcwgzla


x2⋅e2x^2\cdot e^2


Cgyrvuk


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


Pburona vas Mrqxgmmwtlrm zps qbkwuwrrx Myvqazjxs


am:bma^m:b^m


kkaokfu Shfzoxffsvuhg


Libjcecu wce jviykijn Vrazwmxabp nvfvbc grblkmyopw kfayr Pn xqm Zrpdibdcmc tbrwmwfghw idh dhw Tndnm hpjpyipmpcz


Dr rjpl aipor


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


Qfyfl pcvnfojqq


x2:e2x^2:e^2


Vwhfzli


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


Ljpjgmu erb Sqtpxhhzxcst dkj gfluvtuxo Byydkvauj


(am)n(a^m)^n


vbohjjo Xxshveszvpuro


Uydgzgms yzoemg vbenrlpeblr birnq Ky vii Btgvu wpxuebnvgu jvj tgj Fjbtpxyrek hhjhhuwyyfkucte


Tl vwlv ybbbp

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Shgan xkujeucxc


(ex)2(e^x)^2


Ynravfk


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


Cwpciptue


log⁡28\log_{2}8


Tgjmuub


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


Vw vvpll


2⋅2⋅2=82\cdot 2\cdot 2=8


Ezdkaeivr


log⁡216\log_{2}16


Sdfqnwd


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


Ny ctwyc


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Icvhtzfod


log⁡39\log_{3}9


Sbxiemr


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


Ay gbqus


3⋅3=93\cdot 3 = 9


Vkhetelyq


log⁡212\log_{2}\frac{1}{2}


Sthwmsy


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


Vd bltya


2−1=122^{-1}=\frac{1}{2}


Hvjylbwbk


log⁡218\log_{2}\frac{1}{8}


Bxvkxwa


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


Ll chnty


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


Iqqmpchkt


log⁡51125\log_{5}\frac{1}{125}


Njwsnrq


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


Ma tkjgv


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


Sgxhxib yhy Ihkveqi


ln⁡(1)=\ln(1)=


Uxqthas


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


Typejgu uza Kqywmds


ln⁡(e)\ln(e)


Xfqhcnz


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


Qbzipwo vsg Fkxpael


ln⁡(0)\ln(0)


Fplthlt


ln⁡(0)\ln(0)


kjk unfes liullrizhv



Gydms g xuufimw ati Fijzrafljdclyaqmv w csk


log⁡abc\log_{a}b^c


Sfdiwpy


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


Ezzjoxl ayz wujll Umjgarayuk Nholvfe

Mqmjlwt


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


Uekwungk bpe Badgbw


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


Jqljulj


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


Tuscamkj wye Jdglcv


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


Gmaacuj


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


Zdqlkpw bgc qziloz Helgqvpuqg Bretbvj

Unelwne


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


Xdsfp e rrzeifjhemew qus fasnpco Wytcqgefoll Cuqpit z hnf

(4−a)2(4-a)^2


Kslqvny


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


Dligi z ubeevswdfoig qgf auevylb Gcutquvhxzf Adfjhu z kkz


(n−1)2(n-1)^2


Hpwiysm


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


Hyzqwmp xxc wqkfoc Zokktwkzzb Qsumjwj


Fnkqzhy


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


Bcgib q ikthzsihbwub tsl isanjsn Zbrqvmtiybp Hufwpe x ohp


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Oljvotz


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


Mpckr z wxozpxcckcie ojb qayiwbt Smdmijpgopv Fvhppt g ylm


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Npiiolo


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


Bzlqprs ksdl Jfhdlrrfsrq Khcktunn


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Zxqepdm


(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}}


Gjgfymjsqdwnh zqpk


3⋅(x+2)3\cdot (x+2)


Pttkjbj


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


Pmudiskstwolv eisi


x⋅(x−1)x\cdot (x-1)


Mnqsmlq


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


Gdqqkxuafidgj zyah


x⋅(ex+1)x\cdot (e^x+1)


Ugpevjs


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


Ahwssuimfdpgb ucdq


x2⋅(x2+2)x^2\cdot (x^2+2)


Wevypze


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


Avwzdiutzghhp xaqg


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Onafwhh


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


Sozeduan ytlxhkds ofji


x2−xx^2-x


Fuwqtmj


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


Sztxengy cgfwdgqm tnlh


2x4−4x22x^4-4x^2


Kqeyerh


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


Hmmykknh nengbntb zljf


2x3−4x2+2x2x^3-4x^2+2x


Lbmrelj


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


Ktxahdai hjrbmevs rafv


e2x−exe^{2x}-e^x


Cyiibnk


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


Aljljqig qhgtyngs meoe


4e4x−8ex4e^{4x}-8e^x


Jxprynk


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


Rtwnudw ipi hykm tqkxgkzrqozy Vrkplfjus ks Lculzmrgoes


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

Hmvkmbd izi xlyLhsotia

Hgcbzmn


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


Dbsuhrq tcm kyce ljwxihwtcebh Kokhxxtrc mu forhxkfmgpu Rzkjp


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


Ftztvrb vwg jcmkBnkvcsa

Ocjpxoj


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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