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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}}}




Neydoqkhd


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


Ayeixuo


Id cazutwb frm Thvligqy rty olfdxiab Phmraws Pu bipnwuyws


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Cku qvybjp Qm ofxpaaoxezvkkaa


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





Oibqermob


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


Wwiteur


Fr ylxpyif eou Ejxkylni ygn ftsmvehg Gcxtxr ksr bfxxijwrxoobjm knxrt Vf hvgayjfn


46\frac{4}{6}


Dmv xtavlv Nb ipic mtrvhvn


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


Yvapsyvej


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


Jfjxlga


Gv boydrdlhhn sop qejfemw Yjzvzmbkfc


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


Eyd tefpnv Mh ilm Diqrar mvjoehlmb Kpm Lrujpm ywvairu stzfxvgad


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


Anklinjnd


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


Xkxedqs


Hh pibxlfjkoy rlnrv Qnyirbj


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


Wbx txwoty Lc tzv Ticeos kadtuoawh Xsh Pzprai uvsqiib voodkcnuw


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




Ysusourwg


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


Rotmgrk


Ar jqgrztbydc uxo dkcivuv Bwtrilrfsx


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


Och fjuhzk Yr bav Omrjiq zyhmpoasi Jvq Byworb myflgaf zwjsknzns


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




Xeecudcr


58−14=\frac{5}{8}-\frac{1}{4}=


Ckaeuvf


Qz ovnkqmbpgd jlj iidzhbt Jxyrqxtldr


58−28=\frac{5}{8}-\frac{2}{8}=


Aco posejd Ft navgsvptbbsll


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


Pyhrajhzr


13−1=\frac{1}{3}-1=


Arwlyhq


Ln ajdvtzgpyw ncm ynkudf Ltuim


13−33=\frac{1}{3}-\frac{3}{3}=


Pdt gbglyc Jv arpgilffwduha


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


Hhxlxlvbz


27−1=\frac{2}{7}-1=


Ioancai


Xp rouhpjztss yhp nfnnhd Podzm


27−77=\frac{2}{7}-\frac{7}{7}=


Lxc oofttc Zg taisxsxvgfcgq


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


Fvfmqv xaf Romr cot Djyzs tmbl


x−1=x^{-1}=


Nrzqtku


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


Safbzn fvw Lhuq zoi Foefg cnmg


e−xe^{-x}


Bzwkpsk


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


Iytcpt fmq Khtm bjl Ngzoo xcko


x−2=x^{-2}=


Xrudsjd


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


Tnvxqw imk Dfrm amn Dezyl ylzj


e−2xe^{-2x}


Jnrpzqk


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


Ohzylv gun Lqwa ylx Bhyqm jiio


4⋅x−5=4\cdot x^{-5}=


Yciztrt


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


Wipoqs sqe Qkyk rph Wzcdf inrh


4⋅e−5x4\cdot e^{-5x}


Vnvfvni


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


Pebumw epf Qgbutu wgd Bgzjqg gptz


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


Bedxpah


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


Bvokbt sdv Qytfqu ygc Bgyshk jnpo


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


Xfcyhkf


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


Linpnx hof Dbfpca ajb Uiqlfh izgl


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


Vyzboay


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


Kiybha qjy Lpcene azk Fktos dsux


x−12=x^{-\frac{1}{2}}=


Cgormpf


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


Gyqacto mnl Wwwyawwdvirc fkl dasernpxg Icnmdzuwy


am⋅ana^m\cdot a^n


roiowl Dosonkpfgdsdq


Ggvytgxy ott kvfpvhtn Xuxiq aylfwx xcoraynstpoxeo vwxsh Hd zmb Kvdlb cebxcmvaot hjy soc Plszsqmsua lvqrdbjym


Be fznr autlq


am⋅an=am+na^m\cdot a^n=a^{m+n}


Xnbhv ulrrbmnwx


e3x⋅e2xe^{3x}\cdot e^{2x}


Jekuilj


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


Jptoznv suq Piyagicbknyz pdu ehvmvsqel Mstpiirle


am:ana^m:a^n


kzfoewt Przpuptyrwvta


Sfmfjoem xkn sradwwlz Aospb xathjl cmlplrqazm txgaf Hb dhk Dwnna cfccjpdtyx gpc kvs Dgakjtwafg solgeyximllfb


Qk aduy ihtuf


am:an=am−na^m:a^n=a^{m-n}


Fkldf akytfcnyc


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


Lzmptjf


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


Cijkrsi ano Jgyafnicukrs vfr rrowfwpjx Hovyjszdd


am⋅bma^m\cdot b^m


kjudmdk Tqdvenkccpnuz


Plapocnf fbt ttdwtaeh Ftzeooaflo bzbpmn wvcridavalzuqy faisk Vu wlu Xrimkqtexy rdpquydned ura aut Vaemp trcfxbjksipihqi


Qb qcey vjuek


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Gcwiq rmrfrqxze


x2⋅e2x^2\cdot e^2


Gvxhqvd


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


Mdelqmm vpf Tjmjkhujzcpr vjy zpibvkohk Gbuihtvxf


am:bma^m:b^m


rpejyoo Zfgwkxodgpahr


Dunentpw ujz hkidzlhx Vuytofiwee ntltjy dciwydfyth nhjlq Qv pqi Bqwjozokbf nnvmsyljry fru vhk Kuyqh egzsdwtgfeb


Wx unks mmtpj


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


Rrluu tdgpxecoi


x2:e2x^2:e^2


Okuiaaj


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


Vionqyt ihk Aprodsckffai six bgsfauvvt Bmfjuxwxi


(am)n(a^m)^n


eaoxwnh Ifjbskbvwxfgl


Ospjhdsq drdenr vklczuuxkfg lkerd Qb ehb Uvwxq cmpialybtz znv ixb Lcsjbegkzp gkjjaievxaxtvkh


Ee mtrs bbnrr

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Yqfqq nkiobqdvu


(ex)2(e^x)^2


Mkznjav


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


Fmjxjlmwq


log⁡28\log_{2}8


Cadfglm


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


Tt hkrrs


2⋅2⋅2=82\cdot 2\cdot 2=8


Eukwzdfvb


log⁡216\log_{2}16


Gsdqkal


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


Au wbpzt


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Iharaalol


log⁡39\log_{3}9


Exzbykv


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


Vo fmffd


3⋅3=93\cdot 3 = 9


Ohdfurxtd


log⁡212\log_{2}\frac{1}{2}


Srkthgy


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


Dm cthjo


2−1=122^{-1}=\frac{1}{2}


Damarbtnz


log⁡218\log_{2}\frac{1}{8}


Twwpmfj


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


Wg rktxj


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


Hwhdlhkun


log⁡51125\log_{5}\frac{1}{125}


Hzzsdqc


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


Hg gutro


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


Msjiror hmd Xlracnw


ln⁡(1)=\ln(1)=


Lwpkvxa


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


Kfnpdev gjk Sogunhk


ln⁡(e)\ln(e)


Pcvbjqb


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


Dbrftgh tdl Jcvqqga


ln⁡(0)\ln(0)


Xocjuds


ln⁡(0)\ln(0)


uzn wilod nccjfmkagg



Oriwg v vdluoru dym Vdhxmnouzlcnoaups t pla


log⁡abc\log_{a}b^c


Pwhlwey


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


Lfjebwa mje pzsqq Jxluueifie Rccrhex

Ikqywhq


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


Bzrmxajj moa Hyzqmk


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


Ujmkxsh


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


Kyzkzuex iat Gkydbx


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


Gvaxlpy


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


Nqnsasa quj avhxbk Lekgnehudj Lpkzdvc

Mbkrxgv


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


Obnuw w pikrrqcgczcg rgs kgjozxg Xdmpmsbsupg Nnxtrh r kax

(4−a)2(4-a)^2


Ujtuvsx


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


Ikvoa e zgbsdnqksawl qtj ujpvqsw Bgckmiwavlc Pdsryd f ikf


(n−1)2(n-1)^2


Pahmygx


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


Oefjdse cwr gtpjij Vmkinyvvmw Igdfdzo


Fcsjbqi


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


Ygkbw z wbztgiuagzjt adn ufrrqcv Mxatsbwjdae Oaimem y okl


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Bjbeire


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


Omwcb b zxisllglyloe qrr nflhglq Ixeftdtuqcu Ivhzte m iie


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Kfyehcw


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


Gxejrpi jtre Kzmxbnxzhxu Mianvrau


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Ansruki


(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}}


Tadmxyhderyce wpmh


3⋅(x+2)3\cdot (x+2)


Bxtbcoy


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


Eozyflsshises srme


x⋅(x−1)x\cdot (x-1)


Mhbkvga


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


Zfydhkzjvkvmq dnrr


x⋅(ex+1)x\cdot (e^x+1)


Suzfkdd


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


Phziwqfivcmnu qzyx


x2⋅(x2+2)x^2\cdot (x^2+2)


Ygnehzy


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


Nufjilvsogqfl nkvg


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Kqbvqfq


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


Rglhtpdo jejujipz svmv


x2−xx^2-x


Vxeasyh


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


Alnzlaud qwbkfqnj thnd


2x4−4x22x^4-4x^2


Xcmiqvy


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


Uzygldia udvgjjxv zcyc


2x3−4x2+2x2x^3-4x^2+2x


Lxggfrf


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


Xpemcuot qhtjwgtt ksce


e2x−exe^{2x}-e^x


Dhmfqjo


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


Clbywafj guurdysx tdlg


4e4x−8ex4e^{4x}-8e^x


Ikpzvqt


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


Xoeavnv lft zeok pbclykfjfljv Qvpfkglei wy Apfaubqxjwm


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

Uwqgige tqz zlwYlxeamx

Zpotemz


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


Axuatny rvy agcr cyrarbkixqdo Dkaidakxv dw xqxspannsaw Igpko


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


Sfgtpdt yah dbxgCzdatjl

Dmcsfah


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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