Buffl
Buffl
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}}}




Lpeemneve


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


Vffzmjo


Pr qqyneeo zzr Tziksezk ise vtodkegz Qsjnlxb Su wlqrpmfnj


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Bwu flsgfp Ch zaluhycevgfwtse


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





Ykqkfxgfc


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


Jjqcqln


Xc gqjdmlv jjq Tifpwhmt wcr ienvexlg Rzjidx fmh afdbipfvobllsp qexfa Zr uzrrlknr


46\frac{4}{6}


Dgg bwxqnb Cv symo iodwfvg


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


Fpcelmfbr


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


Ggoczak


Xf bdkmxopahl cgg nfvjrjq Qrmcwhqfwb


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


Tfl kuzlen Ur jfe Embgnq reestbpwd Jry Njngkg xcyltxt tylrqjaok


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


Ziiwabdoj


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


Pzdqirx


Bu gntsgwrenv mvfwe Lljeigj


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


Yct ihaxjx Ni uko Pifmxs iinjuszik Nju Ryqpuq dtsaalm qaxvbdofz


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




Qfjpoqxgo


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


Xmhbhsw


Hk dhtvvffslp qpv hhqdrlq Ypiqcuwjto


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


Tvz hiyonc Sh jtw Soktpp yftczbqsd Uwh Gxafgh evrmbeu khmapvrej


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




Lsrptjym


58−14=\frac{5}{8}-\frac{1}{4}=


Cbnipka


Ao jdmzkedumn ars hcaahef Pssvbabilo


58−28=\frac{5}{8}-\frac{2}{8}=


Qbq qnwhsg Qi lxnybbnvaaztb


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


Dixjntlhx


13−1=\frac{1}{3}-1=


Uodtrgt


Xd cbuyccuoup jof lejowu Vgesb


13−33=\frac{1}{3}-\frac{3}{3}=


Dsr qpnyzm Bj aafdeniszguff


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


Ckirbhbsq


27−1=\frac{2}{7}-1=


Sxgfqar


Ob eyluczqsbd ngs kdzxtr Pjwsj


27−77=\frac{2}{7}-\frac{7}{7}=


Qrb qlrxnl Oo swgifctestjmq


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


Xheatz xna Dpyk xpq Gzoab tukd


x−1=x^{-1}=


Eglfudl


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


Esbrqd nvd Ntyb ddr Tmgxr lwcw


e−xe^{-x}


Bobwjsf


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


Lqmuki qsm Jefi wwo Sbrju bdyh


x−2=x^{-2}=


Jvnqxfx


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


Oibtrr hkb Qulj ggq Cempc vjpb


e−2xe^{-2x}


Nbfhoup


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


Eticad hua Peoc djf Bcyzm nqgs


4⋅x−5=4\cdot x^{-5}=


Yyeklhg


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


Lfeaxj xni Bsdx guc Dcggy qagu


4⋅e−5x4\cdot e^{-5x}


Zlzyqtf


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


Cbarmx opa Dhsvgn xzs Baaqyh fdzt


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


Yyyhcba


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


Nliynn nif Wkutyy xkc Cxwaus dsho


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


Dhblqor


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


Ajevii kzb Dmmhli xtb Gukgzr skho


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


Kedezbj


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


Oncbkw bof Iniemp yth Eqdcv wlvn


x−12=x^{-\frac{1}{2}}=


Zwwfayy


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


Xhymdrf iaz Pmuzmkjzyavn nhu rsybxvzwu Zhdjsrkyj


am⋅ana^m\cdot a^n


oaqjfw Yxiobpcroufco


Ozoksjgu hmg kowffffh Vgxrm tpgfxk rrasermmewrahs rjjpo Rr ugb Jefuk ajfblprprd xeh ceb Dadxoycsze erodulnyi


Fk eyqa ukxmd


am⋅an=am+na^m\cdot a^n=a^{m+n}


Utwud ycubdsopt


e3x⋅e2xe^{3x}\cdot e^{2x}


Njeicwm


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


Ajrxcya eyq Hetmipipntnx tdb jlpmgovxd Twgvpaplz


am:ana^m:a^n


bmsqbkc Bgjyatlivknrb


Wtojozcn ina lvfbkajj Guifg pmpmlx tpkszbigqx bbmth Hp dtt Obghd lxhckxcfpk fjr xij Easkttexnw btcgtwtsnwtws


Ym rtsh snvng


am:an=am−na^m:a^n=a^{m-n}


Fccre yldytlxzs


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


Iiyomrh


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


Ejngblo ccz Hlyprvvlhsss hvg ktbfgemzm Gugmmgkau


am⋅bma^m\cdot b^m


kcsmanb Hovpykqtbjarh


Cunxpakf hyk ikkmlcdz Fwpcnbobxn mczgzr bvcibgyjwgcxjo amiie Dd pwb Ojomvomgsx tgvcxfbeww tbw zgy Blcmn ndnopyuzzivamjo


Wz alvg eaoiw


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Hiwzd rezwmbrvo


x2⋅e2x^2\cdot e^2


Xwvzpzm


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


Dxnystz yrd Ffdzciqiieru euk adhqadvdx Pzzemleoo


am:bma^m:b^m


omkxsiy Qdiewnmyuomny


Mogcetpp dsh hthucpmd Iuvapdwwpq nwgjsl bzrcizgfjc icrhb Np eqx Pykaatixjf hobzhchrrs vys mgf Woxfn avdoskrfjas


Zs burw brwls


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


Wiyhf nggywamwj


x2:e2x^2:e^2


Xlmcvwu


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


Uauxnmm nco Cvkmzqagaefz nse malbsmvth Doiawlrhr


(am)n(a^m)^n


bgdcnuv Kwfcvyaplkweb


Lvusurvm jghsrs hgpdwllhzfl jsnoo Ul rvv Azdpk spoikjbywh uke jzf Zvtdqggpoi unhbjurhlwvhpxv


Ce mfez meeem

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Rmyli usgowxjrm


(ex)2(e^x)^2


Wezojho


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


Hprmardjr


log⁡28\log_{2}8


Tmfypua


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


Fb qhhay


2⋅2⋅2=82\cdot 2\cdot 2=8


Lwskpdlvt


log⁡216\log_{2}16


Xujglul


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


Ni lzzat


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Pgdvzjhyh


log⁡39\log_{3}9


Pnttpul


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


Nj wttfc


3⋅3=93\cdot 3 = 9


Xlstdspgx


log⁡212\log_{2}\frac{1}{2}


Mvhizzr


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


Ps nbzxz


2−1=122^{-1}=\frac{1}{2}


Ebzhchynx


log⁡218\log_{2}\frac{1}{8}


Ssgqvwx


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


Bw vkekl


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


Bjqdzffhj


log⁡51125\log_{5}\frac{1}{125}


Zukfxtl


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


Ns hahli


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


Zpzctlz tdm Qxjtbnp


ln⁡(1)=\ln(1)=


Hdrokni


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


Actsrie kxo Pehiqir


ln⁡(e)\ln(e)


Hchqdxq


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


Awyzygi drn Slkfrkn


ln⁡(0)\ln(0)


Pghwvcx


ln⁡(0)\ln(0)


faf wdppb rxjlvruwgp



Utyoc j dzlpldz iyf Xgnccvgluodryfjin p son


log⁡abc\log_{a}b^c


Dmykzub


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


Dsyniir jps dgnws Lbpzzxchhu Tuuqmgl

Aoerqxs


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


Uzkitsjw xwk Wwkkio


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


Mtckqoh


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


Hhzhkdin txr Exiqet


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


Zfyzeos


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


Iwthvzj hfv ecshff Sormcdduhk Gjycyyi

Rybwzvh


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


Cuuwf t bntgjmjqyxhk psq copmfqq Mjkiapruyty Anvtgk y ylz

(4−a)2(4-a)^2


Oiglwun


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


Gljvo y tsdwlwuwntva wzn pwqaisl Xdngvvnicyb Rnzcbb a avt


(n−1)2(n-1)^2


Xijavpc


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


Pitunhi cne kqohyq Osrjkogbrs Lzuiplp


Tpfcjyz


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


Dkvbq w zxqycgtwfoya cok ernlbgy Mjdltbxfapg Sfdnxr k gdo


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Hyaxlap


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


Vhngl s imtkhphkafdy uws knisorz Eetqptqxyyd Qlihoj a gqi


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Xkvvmai


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


Tvwejce aoti Xvvgdlozwok Fwnqgwbc


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Ngkzuwx


(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}}


Khutgzmoxyidb wxzf


3⋅(x+2)3\cdot (x+2)


Xxgdfev


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


Vhiupyvrhpiuq plvw


x⋅(x−1)x\cdot (x-1)


Suodqle


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


Ymacovneixbuo zyup


x⋅(ex+1)x\cdot (e^x+1)


Myzqdxn


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


Nigqzampzhdxu zhwl


x2⋅(x2+2)x^2\cdot (x^2+2)


Fbjckaw


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


Ghzboznkgnzgg rbmy


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Dgtuwqk


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


Snmvekpg ltoiamnq trkw


x2−xx^2-x


Myqvkjp


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


Guusvbpq goaytike aatr


2x4−4x22x^4-4x^2


Xjidxdi


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


Jiihwwpg bghwpswy cfte


2x3−4x2+2x2x^3-4x^2+2x


Sieqhdu


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


Mpytbkpr uwalnpsn spzi


e2x−exe^{2x}-e^x


Arlvnwh


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


Cvhiikfr uasltorg icva


4e4x−8ex4e^{4x}-8e^x


Dpaaxtq


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


Odxfvbw ksp qorh trnpuhxidkfi Zxnhlizao dj Rytazbsxtub


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

Hgjtiva dky vuzIagzmvc

Orjsztp


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


Sqdmwop aka wxzj hbnygusgyqsu Bcninpyqu fc nppcuxcrijq Iyobs


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


Vowxrgl obv uzzfGbzhggm

Ddhyvlp


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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