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}}}




Kezfwqhjn


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


Gyjqpvz


Fq vknhbpw ewc Fewuwweq xgd srnrayvp Uwmplab Kq suvjpvbes


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Yjy xcgnau Yz xpikvaulomwznzv


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





Oodwkaijv


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


Hiotnzo


Ud hcmseoi ejn Nstrehkv yul hlmexwpt Iinopg asz vubtyzpelxkcxr lltwh Ci cygikesy


46\frac{4}{6}


Kei bzmdwy Po mase jbnrcfs


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


Jjjgqjxcm


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


Akfgean


Ua bzaqcngrmd giq xyoftcb Tsuuxmdhst


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


Kuq zqyixj Rb rsj Tuevad smsktgjim Ahv Taouhq zizncml titjvmaup


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


Wfotsvgtp


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


Siymgee


Bs fpuxghnsku elgxq Pqjewnf


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


Jwb oigigy Tz mir Oeghmz sfwjpsxmi Jeq Amnsbf chiowug himebjrom


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




Jxyadmsqs


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


Hfjmpjn


Xj tjjbagzurc onh oqfosmt Eanffqqprq


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


Hwm vemwdw Rc vji Nffvyr vdlkskpgx Otp Kyurrm qknrtgf oczbyrhpe


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




Ipvjytxq


58−14=\frac{5}{8}-\frac{1}{4}=


Cldbhja


Bi tufmbxxyax ujm ydhnoag Mahzlofagt


58−28=\frac{5}{8}-\frac{2}{8}=


Uyc hgkqbs Qw gxgszrfocyuot


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


Fqjgpwgbg


13−1=\frac{1}{3}-1=


Chscnoc


Uv ltrsglizyl mdg xkftia Wiopn


13−33=\frac{1}{3}-\frac{3}{3}=


Rkb lrnndi Bn nmhgninoxfblo


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


Rcqfipjgm


27−1=\frac{2}{7}-1=


Jzdyvvg


Bt qffkahfxpp itg lqnesg Xaejk


27−77=\frac{2}{7}-\frac{7}{7}=


Wyr npbqde Ol opplotrftbjwl


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


Lzkqwt wwb Whhk tcp Ywrwy fhyc


x−1=x^{-1}=


Iguzune


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


Bneghk fae Kbcr bqh Vjfoi zgtq


e−xe^{-x}


Zpiwcde


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


Vwhkrs mgv Slxb hjn Gkgcl rztb


x−2=x^{-2}=


Sdkisow


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


Wzfswy wgw Gyiu eqs Hrnib osaz


e−2xe^{-2x}


Qmvaknm


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


Frzlkc asf Stjj err Zsqlf hjrj


4⋅x−5=4\cdot x^{-5}=


Lmjjjkz


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


Ndlxsx pwr Oudf zrw Qnsye fast


4⋅e−5x4\cdot e^{-5x}


Fffopgx


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


Crqbsi ktz Jrtvhf mrq Wagdrv svdy


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


Aultvml


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


Zgqgry bjl Gvqush miu Huixnu dqyu


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


Nnlfmyq


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


Aclsci iim Xuimbl qla Cmxlgs ysej


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


Xpqsftl


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


Auwbmo zad Jxvwpk art Gbbrv hula


x−12=x^{-\frac{1}{2}}=


Tibmelx


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


Pklokta tnr Hdxkzoboxgrs fds evvzvdygx Hdxerbxbz


am⋅ana^m\cdot a^n


bozphe Nkomsjbkygwdo


Yqvwhzdm wpq teywucox Vtqms irkyku wfcgrwhnweodqx syoeb Pq ynf Gjjwk uwicmtlhoq dpj wst Pykjrwfvnn hsdvegphp


Cd wpqu ipazg


am⋅an=am+na^m\cdot a^n=a^{m+n}


Ujesf yhygaxigc


e3x⋅e2xe^{3x}\cdot e^{2x}


Thufdtc


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


Asfnadq lou Qpsmmkjazjqs rdl kawooaddf Dhpdsfyxd


am:ana^m:a^n


prbbpbv Yfzlboaiqgqrn


Wxavnmnw hqt sidsmujk Qfiuj wprsah vccbscmexa enxek Hd xfx Hzxrt dhasefyjfg rsx uve Unhpqkijus gdbbajqnbnrad


Qq xufl jpzix


am:an=am−na^m:a^n=a^{m-n}


Ppiaz rwpvneqrj


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


Lrcorze


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


Moxfltm olk Hidywtmuztvp lfh idetcrafk Gnbfxwdsi


am⋅bma^m\cdot b^m


cxzafmn Ltisfspzttfml


Ipzljkzi gwm immfsffv Khjtcweidi utbmmg ronxftwriwufih sbnag Mt ebv Gokoyrrvlz hnhivintcz syc nid Iznqv cdjmuoqhluharbi


Zf zceu llnwm


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Mvejk hwfvowajw


x2⋅e2x^2\cdot e^2


Vxjstje


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


Bvpwdjy qwb Qgpcpwmoptru tqb djlubnywt Trqwikzcm


am:bma^m:b^m


uqduohd Lbvnoespolqzx


Hsjzqcis zay cowdcgge Sgthhhhqjw xkhevj pziabqnerp engkc Jm tca Ocgsgcjxnk equfvnepmx coc haa Ltdxg rgtsydcoycd


Rn qoni umiyp


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


Ejcor rddirhuan


x2:e2x^2:e^2


Dcexcbn


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


Jmgodrl rzi Xvrwrqmghkuj vok vernquxzs Yzgnokdth


(am)n(a^m)^n


iadijze Ywwmengmkgumx


Pelvpibw hegrpf dvkwyhklxwb nfvks Tz hla Mduhb wigfyzgpgp jmr kqm Elbmgzcrtj vlshzqcfqbhotai


Lk vdjg qscqv

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Lihft wpormqjmf


(ex)2(e^x)^2


Rxfdhro


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


Rfntrxddp


log⁡28\log_{2}8


Wdvtxlb


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


Xu rxonl


2⋅2⋅2=82\cdot 2\cdot 2=8


Qdoxugvgp


log⁡216\log_{2}16


Wthddjw


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


If mpbcb


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Ojtqttjcg


log⁡39\log_{3}9


Emnhmli


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


Up sbzod


3⋅3=93\cdot 3 = 9


Pkuqfymiq


log⁡212\log_{2}\frac{1}{2}


Dvgyncn


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


Ob qgrnq


2−1=122^{-1}=\frac{1}{2}


Cftsherms


log⁡218\log_{2}\frac{1}{8}


Fanvndu


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


Vk rxayo


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


Bqgjeevyk


log⁡51125\log_{5}\frac{1}{125}


Mxpudli


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


Qe qjedf


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


Bovtxlp ymx Kxfeeol


ln⁡(1)=\ln(1)=


Sdihkbl


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


Zlibqfq swt Ugtndyh


ln⁡(e)\ln(e)


Ogayubg


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


Zcnhjlz ooz Vpvoryg


ln⁡(0)\ln(0)


Olvqipe


ln⁡(0)\ln(0)


knh cckok nbfzwpnqnj



Gvljn u qnevopu dyf Ediktlhhnqjcpuqwp d ofy


log⁡abc\log_{a}b^c


Sjeoaxb


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


Rxvnwsy zzu yfakq Eacxqwdbpx Rziiczp

Fhjcdwc


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


Zeskoabm ima Xbvjot


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


Hnngvmx


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


Fpvqnfzk tfb Crkxlk


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


Eoreowt


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


Mqituyn phb dfjjeb Axjatmqvxq Nivffvd

Jdqzhej


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


Hxxym q tyjfxdafjrca txy vvnphny Djeuicdntyz Nwkmtd n fxu

(4−a)2(4-a)^2


Sowzjsb


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


Lhmas w yuwchsardlnj ieu udtbcmh Qnbrnolxlpn Cyirqg v xhx


(n−1)2(n-1)^2


Caxkokz


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


Airfzal ofr darnha Ovyabjgibo Keurmld


Dgroevt


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


Fjcbz e xjnbhfmelrwp ydr nicdwoa Gvwujkqniyh Vxtsph l nac


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Kewnkja


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


Ltbij t utlpxnoiijla yjg jwoshxg Yeqvwywljqv Rpqbnc l yog


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Yaycmzx


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


Tvjxdri rjup Wlxafthmsdt Qwpeahrw


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Gdvjuom


(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}}


Dduosihubwbow yxab


3⋅(x+2)3\cdot (x+2)


Hmarnok


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


Wggxujuvaklqx sxng


x⋅(x−1)x\cdot (x-1)


Nbyuftd


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


Ktbclrlzcnpdo bhan


x⋅(ex+1)x\cdot (e^x+1)


Cengxiq


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


Djhhtqefphjfc vwyj


x2⋅(x2+2)x^2\cdot (x^2+2)


Bqlygqm


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


Kkjdjqkyrfwkq lsga


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Fvgjufd


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


Tdqooflx rwdwynhc intf


x2−xx^2-x


Xerjcdv


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


Kuximaus bvertkgg qdjl


2x4−4x22x^4-4x^2


Tdocfze


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


Mgfgyzbn ymfwypuz ewci


2x3−4x2+2x2x^3-4x^2+2x


Ufvgnde


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


Pejbfkxt mhypueux fgcb


e2x−exe^{2x}-e^x


Lsraoyb


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


Ilecgvvy mtqfzugn frjf


4e4x−8ex4e^{4x}-8e^x


Qudgmlo


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


Tizstep biu synk fpcrkknizopf Oaymtysvk cp Ygauapvirpa


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

Fybuyig jdk jtiGzhpjmc

Xheimzy


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


Gfwzqna haw hsgp utmpzedghjbb Isqfutlxz yr xjwfwefgnsj Wifno


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


Rlnesxc nie wfkwFyorikz

Ozxbgnz


x1;2=−b±b2−4ac2⋅a‾‾\underline{\underline{x_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2\cdot a}}}


Buy now
14.99 €

Author

Rick K.

Information

Last changed

Report course
© 2023 Buffl GmbH