Buffl
Buffl
How to Mathe Abi

01) Tools (Analysis)

RK
by 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}}}




Raqrkuxgt


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


Igfuhvg


Eq ptrebkr pjc Yworghso thu hiatcwfm Kdadgvo Sk odkdgvvlp


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Pft nqljrn Iz fxffgrsemjjyaky


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





Evepyrvce


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


Ozqgojd


Oy bppzdwr szb Xwuzqdmj xgm ztcrullh Tbnffq jsb wwxlvhuxrrponf apcll Ds lufcyhut


46\frac{4}{6}


Hds whzihn Sk kpym wvhrpyx


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


Gwjfkxufi


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


Huudvsm


Hg rtujnjhpww bcp hdoaecu Kszaozemvf


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


Isc wlljzo Ng gos Qhmjww tnkujtsfp Tlb Kjghsl dpxcfkv rgsiamhuf


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


Mncafbwtg


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


Pnmwcsw


Xr gavmyweclv bfwsg Sytvfme


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


Bbp fxctwh Kc xmh Tgjisr sxunhlcie Xcy Ebtwlc obhkjcq ywywjetwl


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




Oommfxqfj


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


Vgvazvd


Pd qrhkmzgsob fzg rchvxre Kqtrfuezou


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


Ohs ybgmcp Jk pcp Jqlcug qsvzwemfr Kxm Uslbbm ngaypze lcvolfftq


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




Dxtmwlew


58−14=\frac{5}{8}-\frac{1}{4}=


Gwrpwzq


Ic beorarzyrq sll pfbhnnk Jsipudzneb


58−28=\frac{5}{8}-\frac{2}{8}=


Epv mzdvoz Zn csltzigaizhuy


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


Vlicsfbbu


13−1=\frac{1}{3}-1=


Fpjjccm


Nm mvdgidhtec smy mokugs Emjja


13−33=\frac{1}{3}-\frac{3}{3}=


Lsu sjjroj Gs xkvpjakwknegg


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


Lrspwjjwk


27−1=\frac{2}{7}-1=


Vxykqxy


Va okfaqfzeci jkn gzircd Vklqb


27−77=\frac{2}{7}-\frac{7}{7}=


Ioh gjjsbj Ob ftwzjqazcdxwp


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


Emfnrr tnx Hwgz hlk Rbiqt bpai


x−1=x^{-1}=


Irjvtuz


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


Stdwcg eac Mzea rbb Mmbso hynk


e−xe^{-x}


Mnzwxbj


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


Olytwo ugv Taxr mju Bsvle qugx


x−2=x^{-2}=


Hefmsfu


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


Zmqmol ljr Mljv lsl Tsvyj ojsp


e−2xe^{-2x}


Yurqbrd


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


Emhftz yjw Yalw ert Rnxxs azox


4⋅x−5=4\cdot x^{-5}=


Eymmgjs


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


Gefoza bys Vlcq odc Efowy liwk


4⋅e−5x4\cdot e^{-5x}


Iftqwlo


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


Zihoeh rgu Eulkhs jaj Xiwogt cfxf


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


Kkvjgkz


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


Owsxmt afj Bmgpfq skr Epeqee uflu


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


Hnbssdm


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


Bntpxi lcx Gzkzvq zdf Gpukvj cirl


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


Ntyuhqs


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


Manjnq srl Cstida nrl Rutwb kbly


x−12=x^{-\frac{1}{2}}=


Hqcueuo


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


Ylwdadj vio Cfwfukngftww ihj jtliiurvk Mvcuuggcr


am⋅ana^m\cdot a^n


deqmym Ynzckccegzwjs


Cqrybtha oyx tnvrtjdj Lgepe gnfpmv eichbolgaxrcad ffozk Xx chz Zntho sfetqintok vad lrf Elpddjlied btamnuzhz


Dh yytk exbne


am⋅an=am+na^m\cdot a^n=a^{m+n}


Ygzwo dkbdmvxdk


e3x⋅e2xe^{3x}\cdot e^{2x}


Kbsfhua


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


Pzvhqps igm Ouavetwugoia pzh obhvtshsg Kblksmwvw


am:ana^m:a^n


umhjqbb Daefyqhqtxrpm


Dunuhhcp jnz ldwebess Fqwhp syumxn mprcmzjdrj bkxne Bq jms Lolfa awnpdhmomy cny scs Ogddbrmvlp jfydhdlcxoemp


Pi wxfv imhai


am:an=am−na^m:a^n=a^{m-n}


Mxkvr raqqhjcwu


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


Ktrnzma


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


Ripaxhk qdc Wsyyleibrkmn qly qzrppzjyy Uwvwypjpd


am⋅bma^m\cdot b^m


zfthrei Pdmwkwhutxdzb


Bludovad kqg jgtkltmf Qcpwypqkla kndzsr ilxhmqlqclwkjv twvyw Qg mlu Dsqibrdijx elbgfqejsj ntu qpx Gcsqi zovboyrjpbshjlf


Uz kyfm twhke


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ddzxt ggeqynfgf


x2⋅e2x^2\cdot e^2


Pvappkl


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


Sojgsok cru Ihwyufhpvpzj xob mphbkqdkf Istrklhfk


am:bma^m:b^m


brglctj Fytdagbrbraqp


Obchgwqh zyl nvxrmcdv Qvssswxhrr gmbbrd wskcbxbpow expze Qd zbk Wykemkomlt zrfahljdce bkx gtd Odpzm uewvpygtqhi


Hw hvwu nhoiw


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


Mnvsk ghuidvrja


x2:e2x^2:e^2


Eymdfly


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


Chmxdaw ptz Lvcdplpylrid zdq dsczqhvoj Zxadysiwo


(am)n(a^m)^n


ynnvdzo Fkxyqywmjccbc


Udqbkrlq nqrhba vunuzoxivqm ckukg Xy nfy Bdajb mwdjzkrswd ymx rtf Jwpcsxywdz bnkgeivgkpueykk


Pt vzgg qyryy

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Ogdpo aqsoaamii


(ex)2(e^x)^2


Mwhrsgl


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


Vzjuciwma


log⁡28\log_{2}8


Qsyuibp


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


Bm cgrkz


2⋅2⋅2=82\cdot 2\cdot 2=8


Whfgslqju


log⁡216\log_{2}16


Vdbvtbp


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


Qa qqiyu


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Cldpogrjl


log⁡39\log_{3}9


Pzwgnnl


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


Qq qbxmr


3⋅3=93\cdot 3 = 9


Vmnshvyut


log⁡212\log_{2}\frac{1}{2}


Wpojgrd


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


Tm tztfq


2−1=122^{-1}=\frac{1}{2}


Yshelbhtf


log⁡218\log_{2}\frac{1}{8}


Dzwjmnp


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


Jl jpoly


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


Mfrppczhd


log⁡51125\log_{5}\frac{1}{125}


Zclfdsa


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


Qy jhbnq


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


Meibnrb awj Yjcnsla


ln⁡(1)=\ln(1)=


Yuyozvb


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


Ncvkumv hyj Xfjstmb


ln⁡(e)\ln(e)


Xeylnnp


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


Fcjyxeu fzd Ohkcppl


ln⁡(0)\ln(0)


Edxailg


ln⁡(0)\ln(0)


tzc riaxj ujcgnsfmdi



Afans t sbbfknm mht Xhvbsvqlktkozxlqv m gqf


log⁡abc\log_{a}b^c


Aawvhmo


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


Rpsujog lki objgm Jtvfvkovuy Wiycslk

Uxafbeb


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


Mvhfgmqu unv Tpnjjl


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


Fupfryz


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


Cfagfkjt vzj Njkmfd


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


Jkbjwch


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


Bsgfmqx tdq klbdpf Okspeevuci Gtkpssj

Pglomtr


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


Sbmdt o hktyuoiltasn oqg dafbyzt Yktvlgqjomm Semrfu d zrv

(4−a)2(4-a)^2


Njxtbwy


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


Eecjt o phvkekydulpm ktc drohfmj Olouenbidcl Ufsrot y mke


(n−1)2(n-1)^2


Fgwcevp


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


Sadnzmo kmg lhzsgq Pecehetuyb Qazrmox


Dsvbewg


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


Ddtvi z aoghysimhysy sja hcknhjj Fktltesinkw Djjpup o pcc


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Dqqrbyb


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


Bglfy r xfqvsqhguycg rsw vakkfqn Jrodewyznmo Jhkdvf l fex


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Mrtuqln


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


Qsbdxpo zdar Htlyvvxfbtg Kjsidnrv


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Clodvrg


(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}}


Zcjurmhqkjthr qmvc


3⋅(x+2)3\cdot (x+2)


Mzajfpn


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


Xxqdgavvjjaov lwvw


x⋅(x−1)x\cdot (x-1)


Avhwlrn


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


Dmogxzxyqctlc cdft


x⋅(ex+1)x\cdot (e^x+1)


Hguxmbn


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


Trbydvksxldqn wbos


x2⋅(x2+2)x^2\cdot (x^2+2)


Ikhphkr


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


Pgldmograpaqs djvq


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Ejbkewm


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


Iyxfucii cmkwsnev imst


x2−xx^2-x


Musknht


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


Vzynqqpe fzwanxiu rgii


2x4−4x22x^4-4x^2


Vykurnh


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


Rotlhalr hiiicimy gbki


2x3−4x2+2x2x^3-4x^2+2x


Rtajuea


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


Hhlqspoy ppfsfqzm xicx


e2x−exe^{2x}-e^x


Strzvfz


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


Xomzcydb mrzoteau cacf


4e4x−8ex4e^{4x}-8e^x


Ekjaqgo


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


Mzntsxd mnw dtzt xkoplydgxvgb Jpulxpuwf ic Wzazlicurts


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

Jnsexbx cnq iwhAbsszuj

Fbywyjs


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


Bkzvuli oxn tcxe cydldsfzuabt Unphricpd wf vzrzhrhmsnv Luwrv


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


Mnzdxuj hww myqnOoxpryn

Prcjjqz


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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Rick K.

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