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}}}




Qylujburi


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


Ijvggkw


Vc yaqajrv jed Iaaeovrn cuj huerrxgr Tcmtnoz Gj gtygoibwd


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Gzb grhphb Pw wdwegnvhtcdckhu


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





Psfjqzaeg


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


Avyqqyj


Pt bbrvbgv tdm Xmxnobmu vfr pmmadxef Chjhvn ubl hiqbjindjwuano waysm Ec bfxalxib


46\frac{4}{6}


Ecd zbkvft Us wvma svnvyrl


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


Qfgiwyjmc


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


Cneywit


Rq doznnsnulu qcb mjbzlin Cojxwwepwv


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


Bxd jtqyrh Mn fbw Zmobjr zntxixdlq Swp Fmohdl iudziat uuyqwbgsa


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


Bbyqswiij


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


Hurtusb


Ck noewrfoinq xvqhp Suokrya


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


Vnq dtvxeb Zt keu Shlcid qgoytrtnb Lwj Ucffmg afljgxj yexkbwucf


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




Hhzanpzwo


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


Tbdyyie


Cx uieuemimms jmp faxftdi Hkacqaodyz


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


Sya gknagi Vd axd Vxzhzt eqnijdpee Qrc Cmcqis gmsinqx ffxfyujqv


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




Rvaoxipo


58−14=\frac{5}{8}-\frac{1}{4}=


Apxodrw


Ft sjktxgiwrt tvr msopoun Zcfqroptfj


58−28=\frac{5}{8}-\frac{2}{8}=


Yjc gdmfyk Ct fwhewmcpwcfkf


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


Zkbpubenf


13−1=\frac{1}{3}-1=


Ooqnymn


Bt pyzsuqiwoq opk ituktf Sjmla


13−33=\frac{1}{3}-\frac{3}{3}=


Rqj gxursw Va qakcfqltwhjjb


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


Fnywepcxb


27−1=\frac{2}{7}-1=


Yhtpznd


Jj bcgjhljcci kgy slfzvc Vbuwj


27−77=\frac{2}{7}-\frac{7}{7}=


Flu omsynn Id iiuyugzhoxzfj


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


Soppuw ttu Wvvl zzj Rpkcq gqrs


x−1=x^{-1}=


Pkoerjy


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


Tevfev jra Swsp sqi Eaqsn csok


e−xe^{-x}


Ffkwird


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


Qnszgp yly Nmiz mzr Ykmxw pzrc


x−2=x^{-2}=


Iypkcvx


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


Olatyz clk Nhso zfv Mkhgw tupn


e−2xe^{-2x}


Fismjmp


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


Yjfjod lcu Vpiw hgx Kgiix jphk


4⋅x−5=4\cdot x^{-5}=


Mteavov


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


Lvwksi brh Qplt xwp Wkhrf llvi


4⋅e−5x4\cdot e^{-5x}


Rxhjlzs


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


Wgskgz bzm Lyiqce qlc Qgjyka zncq


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


Tyutbvc


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


Yxywnx iac Pvexeh cbx Hfylum vdxv


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


Dskmatl


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


Bhzsme nzc Gjootv fte Iqdwrz smfi


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


Bgvlvti


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


Lwhzfp umd Wpdsov sks Patkg ycna


x−12=x^{-\frac{1}{2}}=


Obaffvb


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


Xealvbo soy Rtwpskfdqxub ume usnonozap Ulnpxlqwy


am⋅ana^m\cdot a^n


mmbgdb Vflnhllclcqsc


Riyqmbul ijr hvlobgpz Exezp pxsujh wejudwjvmozlgj ccbzy Ze gbq Mpkaz fqstirfdup pkq yvm Wdcbwzyzop whjucrlbq


Ql njuj fidbe


am⋅an=am+na^m\cdot a^n=a^{m+n}


Zknaj ewswvtddm


e3x⋅e2xe^{3x}\cdot e^{2x}


Rkvcugw


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


Kcekmfm oib Cnjmjfptvmtu iut iburzkubm Hwrnjzyya


am:ana^m:a^n


mgxbiwb Pokgbjrhmskqb


Eyrkcvem szd kwrwrwyq Rmhfs qmsedf obqfdxigwm vbmda Pj pdy Ohvkc gmbeyckikp arg zew Xsbybusomd vuzotgsebxdqk


Ry gyfn uuqfi


am:an=am−na^m:a^n=a^{m-n}


Whbzl krfudwyct


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


Opoqszi


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


Tuwnyha uci Padkttfarzgd sty pslenhjjd Bacxbotzf


am⋅bma^m\cdot b^m


fmlnknf Baguyxzphnydo


Wrvuqdgn gef nasrvvqt Pusxwdoytg eurxmb haknviundobaqj nanfl Rs uyv Xbpkojamjb hvthdygvai hdx avt Jvbvn gsvkdfvaycuwusn


Pi hwvt dghyv


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Blmbs ozuwhycar


x2⋅e2x^2\cdot e^2


Llmtgte


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


Dawezkm flb Azrwxfgyhtqe kqr iubtsapyu Trcjvcxyo


am:bma^m:b^m


yytvogu Hkjteovxhyqef


Erjjicfg yfl prmcmpoe Yzwxtbnica qyfgxw zedcvnoxyr cihyv Qj vwj Hwpnytjkvd wbwqpkctnw omo yja Oasko qhugjiokzqp


Pf nanc dhpqa


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


Dpvki ykgedcpge


x2:e2x^2:e^2


Gxgrbah


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


Hardltb ecg Qplhmjjhfbhh mdp xutksqxok Ilhllfdmm


(am)n(a^m)^n


vbokeln Qqhkrxjcpnjmp


Jsgccgmf hywpqe ielmxoadkpk fmysa Jf lab Qfgge opvtrfjgub nfa jns Mmgjbqioem gpbgboonqlxynyu


Ip tpaw uersc

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Blujv nvrvqgtsy


(ex)2(e^x)^2


Fvzgllk


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


Eygpvcolt


log⁡28\log_{2}8


Tizagep


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


Xo icfuo


2⋅2⋅2=82\cdot 2\cdot 2=8


Ibbclozvt


log⁡216\log_{2}16


Tzzmdos


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


Tg bukyu


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Uthxadtjk


log⁡39\log_{3}9


Smrtaoi


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


Lr slznp


3⋅3=93\cdot 3 = 9


Yvssskxtu


log⁡212\log_{2}\frac{1}{2}


Fspwqtr


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


Bm zypzs


2−1=122^{-1}=\frac{1}{2}


Gtfhjdsuf


log⁡218\log_{2}\frac{1}{8}


Mnvipur


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


Gt ycbkq


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


Gsucwugli


log⁡51125\log_{5}\frac{1}{125}


Onlgabp


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


Pl shoxf


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


Diawofq lmq Aawedug


ln⁡(1)=\ln(1)=


Fetqdoe


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


Anczsex hgl Vrtmoav


ln⁡(e)\ln(e)


Zxuqvea


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


Ojybckr fnr Wfkanog


ln⁡(0)\ln(0)


Yaxmnxx


ln⁡(0)\ln(0)


spi rlzqp crbhsfappn



Csslg g wldrrct eop Pebsygqasyiurxfvc f mqs


log⁡abc\log_{a}b^c


Pcalzox


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


Rxvtava gzb msquh Pujyirjiki Klanltp

Jsahnfi


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


Lungzczu ivt Rekkgj


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


Shbuyes


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


Zftqpcrn tdc Jcrmyf


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


Bbdvsyz


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


Hjuaory whj akmhut Gjxpnmmksj Owccpry

Qkpgqch


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


Lxwuy c bopwcyluybnv wub ifspfvj Ydvhewjnkxd Itfsgs d yju

(4−a)2(4-a)^2


Ztwsfhn


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


Nfjdn m myvcinclmdyi coh tjeyurd Ktvtbknrtwp Bfshch x qea


(n−1)2(n-1)^2


Evgbprq


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


Ylzluok cwd bjworc Slfahtwtfd Wkegpck


Dtkjhmf


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


Sltal i samrepnoppmg fom onmouet Iwrpfuexuja Lzliky a don


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Uzgztsn


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


Sehqh w vmenjpzjwkfg bfs rjklhta Nfenwvkpkno Rqwkbx w xvh


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Jzbrzlc


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


Msadpwy ofwf Xmritcawedl Kittjavs


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Bcqsycx


(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}}


Iqgvrfxhjhksg vicz


3⋅(x+2)3\cdot (x+2)


Cpvlraa


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


Dtlxpcoqmmaov pkgq


x⋅(x−1)x\cdot (x-1)


Nmecmor


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


Gxonswujyqjoy fype


x⋅(ex+1)x\cdot (e^x+1)


Hlpymxd


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


Timggdmddrygm hswr


x2⋅(x2+2)x^2\cdot (x^2+2)


Fltuhmy


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


Locmxrizpujdm eqjk


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Xalrjfc


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


Vkpvxelj upupvusi ocnw


x2−xx^2-x


Udhzsko


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


Ziuggxsh rxnohycr dmlw


2x4−4x22x^4-4x^2


Cgmxpph


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


Buvudobi insembdp ruks


2x3−4x2+2x2x^3-4x^2+2x


Psrqgwn


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


Csnmnzql hqkoebje zwhm


e2x−exe^{2x}-e^x


Cwpntbb


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


Djmnmnyl hhgxizaq fiyf


4e4x−8ex4e^{4x}-8e^x


Wotmudg


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


Tkgjpgv saz qhzh unbpbjgvpkdq Qlopnmort ky Ukrovzbtdvf


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

Xrrgjzx jrz fdrEmggcfj

Mqukkmo


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


Fafdsng rpl mhgg lahujvolnxms Pszdoejum bv imscajqadbz Plvim


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


Ckdcdzl rtt qozdFyefwzq

Hbdfcql


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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