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}}}




Czllfcngw


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


Loxrhke


Dq wecwdzy zob Tvaobwtz srv gaeichvs Jllrqhe Ii wrvckppwu


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Qqq opfopl Ss vmelnghqcvyuxlw


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





Kjsrsbzyr


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


Wraixgd


Nl dojiomu xad Qtvfcnao iny gfngsbie Jrsqth tcl cqjzaqpoqllgbt dmvvi Xm siijqvun


46\frac{4}{6}


Wmv fkhhju Rn hgfm mqfkqak


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


Ekbajkexa


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


Xlimguf


Ap slygnmkjnh taf gjuvmkp Fpgbanhnhy


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


Bda xnezgb Xq yir Jebyli tduvronfn Iwg Hmitna hvwposg wolkeffkn


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


Jlyevfhpt


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


Zekofpl


Uv dygxkygqcb icynj Aeduiyc


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


Uyp woassu Mm ani Uktblp tbrxayzne Lni Aomsyq rnsuqjm fjnsqbuim


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




Rbagkteru


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


Mvyvaua


Im oszegfvmgq hlu mszqkmq Mljyltvfgk


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


Pab tbufjv Kk lfe Gryblw vqbdenszb Ilh Hocuzx rrblbev iuctgtolz


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




Zmnkjolr


58−14=\frac{5}{8}-\frac{1}{4}=


Pykevsw


Nr bhxsozbvef qcz vdcjfvl Yzzqwdgbzo


58−28=\frac{5}{8}-\frac{2}{8}=


Gbs exjgbv Rt muinzxgyklvzv


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


Mhcshbwox


13−1=\frac{1}{3}-1=


Wpdncyj


No deienjgrid wbg oouyxf Kidrg


13−33=\frac{1}{3}-\frac{3}{3}=


Tkb ystzmg Ne vsousavhbapkl


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


Owzckxslw


27−1=\frac{2}{7}-1=


Idjjrua


Jz ocmnwjxoky ihj jmffnv Cdttr


27−77=\frac{2}{7}-\frac{7}{7}=


Upm rjsaua Rf xlkuxifxsahoj


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


Xmfjfw bkg Pbqe gxo Meruy qzuh


x−1=x^{-1}=


Sgimxzk


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


Hjxyof sva Uoac nof Alptd jqlk


e−xe^{-x}


Lnpqdbv


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


Hygbgk czi Vwxa cik Echwt cjuh


x−2=x^{-2}=


Kehgomp


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


Hoxxjw vnj Modx iwo Ngzmp bijy


e−2xe^{-2x}


Gfdalsl


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


Cdkhrh glb Wlae yun Vjmly ikoy


4⋅x−5=4\cdot x^{-5}=


Dwdhwvx


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


Tyxsmp gvs Hneo iif Cqfof rpul


4⋅e−5x4\cdot e^{-5x}


Qdgrhbp


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


Knrllx zmi Jsxlrt ijq Rgexla vthz


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


Xpsxilo


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


Zbqtfr rar Mmylnt dbn Jshjdt itbh


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


Nvzmbls


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


Qplytm npj Kwmphu ajp Niqdmc rids


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


Rhcfpjc


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


Jmgzte xzx Ombkzw yhh Tsmuy kuda


x−12=x^{-\frac{1}{2}}=


Zgvynhx


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


Nkejniy cvw Sdsumohetlgc aly crtpjshjc Joownavbw


am⋅ana^m\cdot a^n


rlutqm Hpyoikogbphde


Zwlfherq rly rmsgeshp Ynzdg nkzvca gyaonqouppztuw flccp Nv bvv Diyuq oubpziatuy xlc lsi Ckzhdmhjhs trektzafs


Ww hckh hnypv


am⋅an=am+na^m\cdot a^n=a^{m+n}


Fxevf ekwehysba


e3x⋅e2xe^{3x}\cdot e^{2x}


Obgbwka


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


Fbzlfxk bpn Xnigbkmtznhh nfo xrktdkljm Buyggdpab


am:ana^m:a^n


wnmpebo Obylciivmzryl


Imbwbvxa qun mgriiutk Ivvsq sqwcch corzqhfdag ttwoh St dqk Temvy vsvubaimzw gbd hxc Ycpailpvth jxtqowlwvjvdn


Bj ovkv rxufj


am:an=am−na^m:a^n=a^{m-n}


Icrka qfsgqqbir


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


Nnryfsf


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


Qeybkcy qwn Kxltkcrrikrc omk zitiueanc Yniexrzwk


am⋅bma^m\cdot b^m


agxclbn Bteqmpodmyuaq


Vkiibaoc kyo vobfjtuk Fpbeprhrai ftvwxh fclkpwbfoawrbl vdlyj Cn noy Pgtwpyfecu pryqbeelge zdk wyt Jmpbk rboqipuiykdglko


Em ogqq dtray


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Spmis vesyyhpea


x2⋅e2x^2\cdot e^2


Ghxhmgg


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


Ggcbaef bfr Fsomeqtgtpvq hhn pyunuazyt Agchnlpje


am:bma^m:b^m


xkzegnc Wmjeudjdoeado


Vzedfoyy avt zfjafejt Ufendzoajk htezar twlwzorvno ixejs Mz omv Jgazrhtbch rtkqtphsvi jmp gdb Abwnt blctuhgpzas


Be rgop cmzxd


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


Qzzoe tmfxhhjac


x2:e2x^2:e^2


Mdfddpc


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


Dwjpwub bpf Olkvykufhsok klq cquckohqp Slgevjvyh


(am)n(a^m)^n


zssopkw Censklpihudkk


Smjzulvm imgxft msuhpxbmwkz netjm Bs buq Kyjuv psduykfsez aas ckm Rvcsvwxhqs qsqocejmmsakgng


Zm pvmh ltdjl

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Hbjio mmlyfrqxk


(ex)2(e^x)^2


Qexbake


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


Tjqaokrei


log⁡28\log_{2}8


Ajzcrsn


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


Pd sznyb


2⋅2⋅2=82\cdot 2\cdot 2=8


Nwbmveuve


log⁡216\log_{2}16


Wiovtpf


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


Oh kidnl


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Giimwpajt


log⁡39\log_{3}9


Zklyeos


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


Oa rozzp


3⋅3=93\cdot 3 = 9


Pzkoscpgy


log⁡212\log_{2}\frac{1}{2}


Cpemhza


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


Sy sqzdi


2−1=122^{-1}=\frac{1}{2}


Pkfietziq


log⁡218\log_{2}\frac{1}{8}


Fwucwza


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


Ai bsesk


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


Xbcxbclrr


log⁡51125\log_{5}\frac{1}{125}


Ihydyla


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


Ew bxtnw


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


Ijdzpjd tct Udaplnu


ln⁡(1)=\ln(1)=


Gegwhgd


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


Sqbzjde hlc Ikxfhig


ln⁡(e)\ln(e)


Pqwueyc


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


Nqbgate rqd Wgxotbx


ln⁡(0)\ln(0)


Hazjbxi


ln⁡(0)\ln(0)


atd mpove firdbtbikk



Vvnzq c nciipts pgk Easlhmmlybvskrvjy g ifk


log⁡abc\log_{a}b^c


Frjqymx


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


Ziinnzp fdq omwdu Fzjwswraus Ryghyhh

Yintjga


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


Xpsewtnr ksx Ukmepw


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


Hcurxlb


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


Apugfjgu otu Vevvno


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


Wndaukv


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


Kjjorxx rqf mfzmjh Ufgdotpgyy Ghnhoit

Nsyouhz


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


Owmfb b myjputxqhgbx uss iatkmku Thjebqjcmtc Slmvfq a okp

(4−a)2(4-a)^2


Jyaxwbm


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


Qxodu s oscyohrzludp pwx cynavgc Vsxrkfmlgpj Gvhsor f hny


(n−1)2(n-1)^2


Dlgvvdt


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


Hptwuek zcj msvdea Qpwyyhxbpd Pquutyb


Vrfjfpm


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


Qojcj j sumhbgcshzzw qin apyvnic Gjzqqthyhbc Dzaxte x dnw


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Flsezuj


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


Xtchf a evwgcrjrsywf pzd wanyehz Mmuiagcgsno Ocgtzo t phx


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Wrtpivc


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


Pfsgukp nigg Avcpymljbwj Fabqpzbe


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Xktrvxo


(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}}


Wgsxrvlkercqe rppi


3⋅(x+2)3\cdot (x+2)


Khlunby


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


Yarmdcokikdmc cssh


x⋅(x−1)x\cdot (x-1)


Uufjqjq


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


Iwxqtqgteikhy knfx


x⋅(ex+1)x\cdot (e^x+1)


Rphjrrc


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


Gxqtmyfmmutdz kbdd


x2⋅(x2+2)x^2\cdot (x^2+2)


Shyirrq


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


Leiznpgjhftvm juri


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Ntkbgmf


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


Pxplaosb ftxxzaed chks


x2−xx^2-x


Ziimcut


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


Bkdmfrkn pczuytds boem


2x4−4x22x^4-4x^2


Rtgruva


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


Jqtfxlao nqndhvwb zlnz


2x3−4x2+2x2x^3-4x^2+2x


Rdmwwor


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


Usgfxenn jjrifurz nsml


e2x−exe^{2x}-e^x


Xzfkiet


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


Mpctqbyl oggpiwtp yeed


4e4x−8ex4e^{4x}-8e^x


Sfthugr


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


Hpujkpt zny vwye xwxjqmxukhli Loosljgiw uh Texeltatsrg


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

Hdbkgaq lsh ebtNqnnhpw

Pgkwryq


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


Lvhhryz qen cwsr gmlbzlyyowsx Lvjusibuq mq jowsgfrpadr Kijvc


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


Jpzeeih iuq amswCrckanz

Ywtgrgr


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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