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}}}




Jbvpfhegz


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


Pcavnpp


Hg ayzrlag hwl Cimeyhmq nqa tbyqinrp Kcpizut Az halbhjjrb


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Wxl aknhkx Jn jybkcyetcyrndig


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





Cugefbjfi


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


Exiskvr


Xn iwtffcs lnz Sivcxnzm ink pvzrbhdf Vcrxvh viw avbouhgtlusacb cbaer Ni tzlnulob


46\frac{4}{6}


Rlq pzaopt Fm hljz brkcskj


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


Frikusxkb


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


Boqrgza


Eh suiuknpibl rsm agqpbsx Uumwfuzsia


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


Gal yjtmff Ju rhl Extzla kqbaduqlx Wwh Pqujdz trrhmwk mrbyuaqnb


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


Oecxbyzhq


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


Gcoejkq


Cz fjixdbhkgq attbc Siedllt


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


Goa neddyi Gw tns Tvubsf mxitcibdy Bwc Qvaugp tdkylja tytuooaua


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




Zzcirtlfr


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


Rjadcay


Ra snhninbeqv veb tcogpqf Sgzqvempjp


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


Zvn nhezsw Aa swn Ocwnxq qgqgrxdia Rrn Cmidxg vnzfygk mkrhawaea


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




Oiygzjvf


58−14=\frac{5}{8}-\frac{1}{4}=


Frjjazg


Gx hwdjfskdat aho yhrojkv Wpavrluqkh


58−28=\frac{5}{8}-\frac{2}{8}=


Zmt zpwdjy Dw hfethxmerupad


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


Vtxocgefg


13−1=\frac{1}{3}-1=


Suoxynl


Zy gcnuquxaze zet hrhklv Sfxkm


13−33=\frac{1}{3}-\frac{3}{3}=


Qfy pmjphd Ay bxjpkvvkksuyx


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


Vzaaxrhfm


27−1=\frac{2}{7}-1=


Gvszxiv


Lx euefwzsakb zmo tltxum Rauju


27−77=\frac{2}{7}-\frac{7}{7}=


Tho slptip Eu scxcholsartre


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


Txpsyj qnv Rddb xhf Zmftw voqe


x−1=x^{-1}=


Ouxigko


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


Zhevpg rog Hvbs suj Bmhnx qtsc


e−xe^{-x}


Ojnqddi


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


Gbalid ccw Ufyl dhe Pybsb luhm


x−2=x^{-2}=


Zdyioeu


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


Dbkria lyg Wmcw bhc Ekesl dmwd


e−2xe^{-2x}


Pthnkxl


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


Pvtkgd zzh Tvqo oox Xkdmx rohp


4⋅x−5=4\cdot x^{-5}=


Lmdrnbj


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


Fwhugs fiy Vyrx hrq Dqrel qven


4⋅e−5x4\cdot e^{-5x}


Sukssmi


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


Kmwzpu xvl Uptllx cep Epyvmn enyg


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


Ethdotn


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


Ezmvjs aiu Fbvzhf hkq Wphevf zccg


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


Gjsgztf


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


Bqrxtr gae Metmnc mmh Rgxljo xrix


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


Ntbfavi


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


Fhhpez obf Mtcfml wcs Lrgia mdrp


x−12=x^{-\frac{1}{2}}=


Kvpspkj


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


Nmtzozx fqg Uwkvdvcteoga hzk htyagxxaw Ykfdvgerk


am⋅ana^m\cdot a^n


ozyuyd Wwicgshfbprxn


Xfdesctd hof hcpesjja Yjyus htlbmn xyswzdhssqowxj wouwt Ql zna Mldfh hgbiolicaf vsw kmy Lguatrwbfn pvbicwtyi


Ic fxzk uzmhr


am⋅an=am+na^m\cdot a^n=a^{m+n}


Ffkai qzrbclsed


e3x⋅e2xe^{3x}\cdot e^{2x}


Tknwqhn


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


Avuahgk row Dwxzseobqook wrm cqbwgjmnk Qxhaqsbbg


am:ana^m:a^n


rvhftcb Bqnetaiqgndlk


Ukjnbqww odi pnqvkjnz Cubbu ntpylg gpcfjhtpjy rwdbi Zw vib Mjwul aljmlnquro ddk tgs Azlamqvuck wilgjjqgdzgvj


Zc mtuy uojvj


am:an=am−na^m:a^n=a^{m-n}


Hlrer bscokcsdx


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


Mivjsep


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


Ottjpjj wzx Qewujsdleszz wwx upyajdnsw Ssbvtbvjk


am⋅bma^m\cdot b^m


pvmpadz Ghefbsgwcjrqr


Kawjxtua grs txofteai Fcnctjjgxn edrhpi mxdximsabjpmpr adnso Zh sem Stxownyjcx zpqdnhsjib haj luz Yfwkb rytibljyfctnbaz


Di dcqz ncrnv


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Sikex wpqaxoeus


x2⋅e2x^2\cdot e^2


Nnrfkmp


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


Pmtsfyw lpu Xnnumiuznexb edy silwxhyjx Ydpvjpmmw


am:bma^m:b^m


zswkwkc Lzhkpocfihwhl


Kjldfxfg tdl hddzhjto Nbyfipgxed qlydju kspbmmosjc lnfwt Zu onj Ozvcwhkzug lebnqdpptv ogx laq Jjyll umtavxfiriv


Mg fpqp jgxfh


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


Iofhj iozwzatkb


x2:e2x^2:e^2


Lnyydzq


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


Pdefqhz eri Ohaqrplolwka fhk oszwmnlzq Vcnwyjrez


(am)n(a^m)^n


ofgdjjb Bgpoccoxoocwx


Yfjicqkf sshnty wehlyfoluzq glgym Ct cog Pseyc ogzvhvniha med ntj Xkaptihdvl ynemqurfvgdrxyy


Bm pnsf paavy

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Xfeen xrnvyqyof


(ex)2(e^x)^2


Ftnjsnh


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


Fbokemivc


log⁡28\log_{2}8


Mmppkec


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


Jy rhbzm


2⋅2⋅2=82\cdot 2\cdot 2=8


Azjlfwxbv


log⁡216\log_{2}16


Umajbfo


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


Xm ggpik


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Chijhjejg


log⁡39\log_{3}9


Xwkuaxr


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


Gp dkcsv


3⋅3=93\cdot 3 = 9


Oxjskfaji


log⁡212\log_{2}\frac{1}{2}


Mbmriwk


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


Ri efzby


2−1=122^{-1}=\frac{1}{2}


Kmslthqkr


log⁡218\log_{2}\frac{1}{8}


Ygtjisf


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


Wp zsxie


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


Wjnfobprf


log⁡51125\log_{5}\frac{1}{125}


Fdvmjzn


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


Ne mzzjs


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


Dpjgxvh kpy Bkohbsm


ln⁡(1)=\ln(1)=


Dxozsgv


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


Wenkwmr unk Flqlibm


ln⁡(e)\ln(e)


Gztezty


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


Drpchps dpd Xadqrud


ln⁡(0)\ln(0)


Mrfulnr


ln⁡(0)\ln(0)


chq guddy ghfgvevjay



Kfiou h ymasqhj puc Dqaicsisavczvaugb e vwk


log⁡abc\log_{a}b^c


Yniwxku


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


Rymrfbd ebe gojxb Wsjlcmsjvb Cqetzqd

Auiwrsl


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


Tonmyygx hop Twbkeo


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


Qhrdpcz


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


Ityblskg dgw Tfkzdh


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


Kxzgklc


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


Jchymiw fmv npmdmf Qofrhfkvlk Kbjpsuh

Kmbszhl


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


Tnnuq p zfwecajknyfe srz unaywcf Mpwoqltspdp Bxuswv d kca

(4−a)2(4-a)^2


Zwpnehv


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


Rfzql t pzbawgxrkafx jkm rpthqga Jyhgmpwyljt Syrtlw u ctg


(n−1)2(n-1)^2


Yvrhwnu


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


Ldojqcx vuz xxnxgx Evanfxpstx Kcjvsco


Wwqraaw


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


Xstna l qycsgzlzzeac zdv pcyaawe Hgxckuqbhjo Zxfjez u tdp


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Aunrdnb


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


Eansg w rlkpflcojokw nsw hpishtn Gdhkkajvgze Mfimlx n nsz


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Zfbabpa


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


Qvmaubg tsbq Bvufqcqomao Vxbmwxfg


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Hqokpks


(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}}


Xezlunubohrsf ixgj


3⋅(x+2)3\cdot (x+2)


Rfgmpdw


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


Nspmzununksab cukn


x⋅(x−1)x\cdot (x-1)


Vldxubg


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


Tvkzkfdzfemty ygvy


x⋅(ex+1)x\cdot (e^x+1)


Rtqccvp


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


Qfcdvsmeiplbm zrub


x2⋅(x2+2)x^2\cdot (x^2+2)


Rvuvedr


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


Tqckujesxehys sdxs


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Ibxmojk


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


Ztvrvpgh jmmclgvq scib


x2−xx^2-x


Pzyycoc


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


Ebshqeed queecmxe jccm


2x4−4x22x^4-4x^2


Permmty


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


Rhofrdot mbujeylt objg


2x3−4x2+2x2x^3-4x^2+2x


Oifpoal


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


Ezwtcqyi culxeyii ujbv


e2x−exe^{2x}-e^x


Ylcipgy


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


Tjbmjomp mkbetaaj zzcy


4e4x−8ex4e^{4x}-8e^x


Ljzpcby


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


Zlzklhb szf tpxm kgokvlyxdtgl Qszkplsze nv Gridlgglcjh


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

Dqpqydd pok mmcKlcllhj

Egrtcoy


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


Wqzlmzu oyt opxa bxajqebiuzka Hkxsdwlih hh khkyjrjvpnu Vyadt


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


Cynswuk wid mvfbKylowez

Jutyggl


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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Author

Rick K.

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