Iingxaki ekufuneka isonjululwe liSebe nxaki. Ngesisombululo iingxaki zezibalo

Ekuhambeni yesikolo wemathematika ofunwa ukuhlangabezana neenjongo. Ezinye bazithambisa ngendlela amanyathelo ezimbalwa, abanye ifuna puzzle ethile.

Iingxaki ekufuneka isonjululwe liSebe lenxaki, kuphela ekuqaleni emehlweni nzima. Ukuba uqhelisela, le nkqubo iya oluzenzekelayo.

iimilo zejiyometri

Ukuze siqonde lo mbuzo, kufuneka ukuya core. Ngocoselelo zicacelwe yintsingiselo imeko, kungcono ukuba ukuphinda-funda izihlandlo eziliqela. Imingeni kuba equation kuphela xa uqala nzima. Xiya xikombiso lexi ukuqala i ilula.

Dan uxande, kubalulekile ukufumana indawo yayo. Ngenxa: ububanzi kwi-48% ngaphantsi kwe ubude kujikelezo liyi iisentimitha 7.6.

Ukusombulula iingxaki zezibalo kufuna ngenyameko vchityvaniya, logic. Sisonke, makhe ukujamelana nayo. Oko kufuneka kuqala zonke ukuqwalasela? Thina ukuchaza ubude x. Ngoko ke, kule lenxaki, ububanzi iya kuba 0,52h. Sinikwa iperimitha - 7.6 isentimitha. Sifumanisa semiperimeter, oku 7.6 cm umahlule ngo-2, nto ilingana ngokweesentimitha 3.8. Sikuphathele inxaki apho sifumana ubude nobubanzi:

0,52h + x = 3.8.

Xa sifumana 'x' (ubude), kulula ukufumana kunye 0,52h (ububanzi). Ukuba niyazazi ezi amaxabiso amabini, sifumana impendulo yombuzo engundoqo.

Iingxaki ekufuneka isonjululwe lenxaki, ayikho nje kunzima njengoko ibonakala, ukuba ukuziqonda kumzekelo wokuqala. Siyifumene ubude x = 2.5 cm, ububanzi (y oboznchim) 0,52h = 1.3 cm. Yisa le ndawo. Yeyona S indlela elula = x * y (ukuba oonxantathu). Xa ingxaki yethu S = 3,25. Oku kuya kuba impendulo.

Makhe sijonge imizekelo yokusombulula iingxaki ekufumaneni isithuba. Kwaye eli xesha, sithatha uxande. Ngesisombululo iingxaki kwimathematika ekufumaneni iperimitha, indawo, amanani ezahlukeneyo kaninzi. Sifunda amazwi ingxaki: anikwe uxande, ubude bayo ububanzi 3.6 iisentimitha ezingakumbi, nto leyo 1/7 yomjikelezo mzobo. Fumana ummandla uxande.

Kuya kuba lula ukuba Zimiseleni ububanzi variable x, kunye nobude (x + 3.6) cm. Sifumana lokubiyela:

P = 2 + 3.6.

Asinako ukusombulula inxaki, kuba oko variables ezimbini. Ngoko ke, sikhangela kwakhona imeko. Ithi ububanzi ilingana 1/7 yomjikelezo. Sifumana i equation:

1/7 (2 + 3,6) = x.

Kuba lula yesisombululo, phinda kwicala ngalinye lenxaki ngo-7, ngoko lahla iqhezu:

2 + 3.6 = '7x'.

Emva kokuba sifumane izisombululo x (ububanzi) = 0.72 cm. Ukwazi ububanzi, nobude yokufumana:

0,72 + 3,6 = 4,32 cm.

Ngoku siyazi ubude nobubanzi ehambelana kumbuzo engundoqo yintoni na indawo yoxande.

S = x * y, S = 3,1104 cm.

Iitoti yobisi

iingxaki Ukucombulula ngokusebenzisa i equation ebangela kobunzima obuninzi esikolweni, nangona ukuba lo mba iqala kwibakala lesine. Kukho imizekelo emininzi siye kuqwalaselwa ngokumiselwa iindawo amanani, ngoku digress kancinane geometry. Makhe sibone ukuba umsebenzi elula yokulungiselela iitafile, zisiza abangaboniyo: njengoko data ukunceda ekusombululeni ebonakalayo ngakumbi.

Cela abantwana ukuba bafunde imeko ingxaki nokuyila itshati ukunceda ebhala i equation. Yiloo meko: kukho iinkonkxa ezimbini, ubisi lokuqala izihlandlo ezithathu ngaphezu elinesibini. Ukuba wokuqala kuphalala iilitha ezintlanu elinesibini, ubisi kwahlulwa ngokulinganayo. Umbuzo: zingaphi iitoti yobisi nganye?

Ukunceda ukusombulula imfuneko ukudala itafile. Njani ukuba ijongeke njani?

isinqumo

	kwaba	kwaya
1 ungakwazi	3	3 - 5
2 iinkonkxa	x	x + 5

njani le uncedo ekuqulunqweni lenxaki? Siyazi ukuba ngenxa yoko ubisi wayelingana, inxaki iya kuba ngolu hlobo lulandelayo:

3 - 5 + x = 5;

2 = 10;

x = 5.

Siyifumene ukwenza isixa sokuqala ludala ubisi ku yesibini, ngoko ke lokuqala: 5 iilitha * 3 = 15 zobisi.

Ngoku, enye ingcaciso encinane phezu kwetafile umzobo.

Kutheni na wokuqala itoti ebhalwe 3: kwi imeko watsho phandle ukuba ubisi izihlandlo ezithathu ngaphantsi kwe kwi Madikane yesibini. Emva koko sifunda ukuba iilitha zokuqala 5 iinkonkxa avuza, ngoko yaba 3 - 5, kwaye eyesibini wathulula: x + 5. Kutheni sibeka i kophawu olulinganayo phakathi yeekota ezimbini? Imiqathango ingxaki ithi ukuba ubisi iye ngokulinganayo.

Ngoko ke sifumana impendulo: kuqala itoti - iilitha-15, kwaye eyesibini - iilitha ezi-5 zobisi.

Ukumiselwa kwe ubunzulu

Ngokutsho ingxaki: ubunzulu kakuhle lokuqala kweemitha ezi-3.4 ngaphezu owesibini. Iqula yokuqala yanyuswa nge leemitha 21.6, kwaye eyesibini - izihlandlo ezithathu, emveni kwezi amaqula izenzo ubunzulu efanayo. Kufuneka ukubala ukuba ubunzulu kakuhle ngasinye ekuqaleni.

Iindlela iingxaki zokusombulula baninzi, kunokwenziwa yi isenzo okumisele zibalo okanye inkqubo yabo, kodwa ukhetho lwesibini kakhulu lula. Ukuya itafile isigqibo sotavim, njengoko bekunjalo kumzekelo wokuqala.

isinqumo

	kwaba	kwaya
1 kakuhle	+ 3.4 x	x + 3.4 + 21.6
2 kanye	x	3

Sichubeka kumalungiselelo lenxaki. Ekubeni ubunzulu kakuhle baba enye, iye le fomu ilandelayo:

x + 3.4 + 21.6 = 3;

x - 3 = -25;

-2x = -25;

x = -25 / -2;

x = 12.5

Siyifumene kubunzulu lokuqala kwiqula yesibini, ngoku ziflaya zokuqala:

12.5 + 3.4 = 15.9 m.

Emva kokuba izenzo zirekhodwa impendulo: 15.9 m, 12.5 m.

abazalwana ababini

Qaphela ukuba le ngxaki yahlukile bonke abo ngenxa yemeko ophelileyo ekuqaleni inani elifanayo lezinto. Ngako oko, le theyibhile abangabancedani senziwe ngendlela reverse, okt, ukususela "waba" a "iye".

Imeko: abazalwana ababini wanika amandongomane ngokulinganayo, kodwa umdala uHanani umzalwana wakhe omncinane-10, emva koko lenci nati izihlandlo ezihlanu ngaphezulu. Zingaphi zepistasi ngoku wonke umntwana?

isinqumo

	kwaba	kwaya
eziphezulu	x + 10	x
abancinane	5x - 10	5x

Kulingana:

x = 10 + 5x - 10;

-4h = -20;

x = 5 - nuts ungumzalwana wakhe omdala;

5 * 5 = 25 - umzalwana omncinane.

Ngoku bhala impendulo: 5 nati; 25 nuts.

ukuthenga

Isikolo kufuneka ukuthenga iincwadi zokubhalela, owokuqala wesibini ebiza kakhulu kwi 4.8 engange. Kufuneka ukuba ubale ukuba ingakanani na incwadi enye incwadi mnye, ukuba kuthengwe iincwadi ezingamashumi amabini anesihlanu yamanqaku enye uhlawule isixa enye imali.

Ngaphambi kokuqhubeka yesisombululo, kuyimfuneko ukuba uphendule le mibuzo ilandelayo:

• Yintoni na le ngxaki?
• Ingaba uhlawule malini?
• Yintoni ukuthenga?
• Ziziphi milinganiselo equalized kunye nabanye?
• Into ekufuneka uyazi?
• Yintoni ixabiso ethathwa ukuba x?

Ukuba uphendule yonke imibuzo, badlule isigqibo. Kulo mzekelo, njengoko ixabiso x yamkelwe ixabiso nencwadi, yaye iindleko zeencwadi. Cinga iinketho ezimbini kunokwenzeka:

1. x - ixabiso yokubhala, ngoko x + 4.8 - ixabiso le ncwadi. Ngokusekelwe kule, siye sifumane equation: 5 = 21x (x + 4.8).
2. x - ixabiso le ncwadi, ngoko x - eziphathwayo ixabiso - 4.8. Inxaki unalo fom: 21 (x - 4.8) = 5x.

Ungakhetha kuba baziqwebela khetho lula ngakumbi, ngoko thina ukusombulula ezi zibalo ezimbini uze uthelekise iimpendulo, ngenxa yoko, kufuneka babe enye.

Indlela yokuqala

Isisombululo lenxaki lokuqala:

5 = 21x (x + 4.8);

4,2h = x + 4.8;

4,2h - x = 4.8;

3.2x = 4.8;

x = 1.5 (engange) - ixabiso yokubhalela mnye;

4.8 + 1.5 = 6.3 (engange) - nendleko incwadi eyodwa.

Enye indlela yokusombulula le inxaki (ipharenthensisi yokuvula):

5 = 21x (x + 4.8);

21x = 5x + 24;

16x = 24;

x = 1.5 (engange) - ixabiso yokubhalela mnye;

1.5 + 4.8 = 6.3 (engange) - nendleko incwadi eyodwa.

Indlela yesibini

5x 21 = (x - 4.8);

5x = 21x - 100,8;

16x = 100,8;

x = 6.3 (engange) - ixabiso 1 incwadi;

6.3 - 4.8 = 1.5 (engange) - indleko incwadi yokubhalela.

Njengoko kubonakala kule mizekelo, iimpendulo ziyafana, ngoko ke, le ngxaki isonjululwe ngokuchanekileyo. Qaphela ukuba isigqibo esifanelekileyo, kumzekelo wethu akanaye impendulo negative.

Kukho nezinye iingxaki ekufuneka zisonjululwe ngoncedo lenxaki, ezifana intshukumo. Cinga ngakumbi kule mizekelo ilandelayo.

ezimotweni

Kweli candelo siya kugxininisa imisebenzi motion. Ukuze ukwazi uzisombulule, kufuneka wazi umthetho ilandelayo:

S = V * T,

S - umgama, V - uqina, T - ixesha.

Makhe sihlolisise umzekelo.

imoto ezimbini washiya ngaxeshanye ukusuka kwindawo A ukuze abonise B. Umgama lokuqala uwonke ohanjwe ngesantya esifanayo, kwisiqingatha sokuqala indlela yesibini kuphephetheka ngesantya kwe-24 km / h, kwaye eyesibini - 16 km / h. Kuyimfuneko ukuba ukumisela isantya sithuthi okokuqala ukuba akhombe B xa beza ngexesha elifanayo.

Yintoni ekufuneka zengqokelela lenxaki thina; variable V engundoqo ₁ (nesantya imoto yokuqala), minor: S - indlela _T-1 - okokuqala kwemoto endleleni. Equation: S = _V-1 * T _1.

Ngokubhekele phaya: kwisiqingatha sokuqala indlela isithuthi yesibini (S / 2) waqhuba ngesantya V ₂ = 24 km / h. Thina ukufumana athi: S / 24 * 2 = T _2.

Inxalenye elandelayo indlela lwahamba ngesantya V ₃ = 16 km / h. Thina bafumane S / 2 = 16 * T _3.

Ngaphezu koko ke kubonakala imeko ukuba izithuthi kufika ngexesha elinye, oko T ₁ = T ₂ + T _3. Ngoku kufuneka ukuvakalisa variable T _1, T _2, T ₃ iimeko zethu zangaphambili. Thina ukufumana inxaki: S / V ₁ = (S / 48) + (S / 32).

S ukwamkela yunithi kunye ukusombulula inxaki:

1 / V ₁ = _1/48 + _1/32;

1 / V _{1 = (2/96) + (3/96} ) ;

1 / V ₁ = 5/96;

V ₁ = 96/5;

V ₁ = 19.2 km / h.

Lo impendulo. Iingxaki ekufuneka isonjululwe lenxaki, nzima xa efika kuqala. Ukongeza kule ngxaki kubonisiwe ngentla-uyakwazi ukuhlangabezana ukusebenza, koko zichazwa kwicandelo elilandelayo.

umsebenzi umsebenzi

Ukusombulula le hlobo lomsebenzi kufuneka wazi ifomula:

A = VT,

apho A - lo msebenzi, V - imveliso.

Ukuze ufumane inkcazelo eneenkcukacha kakhulu imfuneko ukunika umzekelo. Subject "Sombulula equation Ingxaki" (grade 6) abanakuzeyisa iingxaki ezinjalo, ekubeni kwinqanaba nzima ngakumbi, kodwa kunjalo unike umzekelo ngereferensi.

Fundisisa ngokweekota: sisebenzisane abasebenzi ezimbini ukwenza isicwangciso iintsuku ezilishumi. Kufuneka ukuba ithatha ixesha elingakanani umsebenzi wokuqala ukuba enze imigaqo efanayo ngokwabo. Yinto eyaziwayo ukuba enze iintsuku ezimbini imali umsebenzi njengoko umntu wesibini ngeentsuku ezintathu.

Sombulula iingxaki behlanganisa kwizibalo ufuna iimeko ngenyameko ukufunda. Into yokuqala safumanisa kwakulenxaki ukuba umsebenzi abakacaci, ngoko njengeyunithi makayithabathe, oko kukuthi, A = 1. Ukuba ingxaki libekisa kwinani elithile iindawo, okanye iilitha, umsebenzi kufuneka athathe le datha.

Thina ukuchaza olukhupha imveliso kwindawo yokuqala neyesibini olusebenza _V-1 kunye V _2, ngokulandelelana, ngeli xesha, mhlawumbi barhola equation ilandelayo:

1 = 12 (V ₁ + V _2).

Yintoni le equation isixelela? Ukuba wonke lo msebenzi wenziwa ngabantu ababini kwiiyure ezilishumi elinambini.

Emva koko siya kuthi: 2V ₁ = 3V _2. Ngenxa yokuba lowo wokuqala wenza kangangoko yesibini kwabathathu iintsuku ezimbini. Sibe nenkqubo kwizibalo:

12 1 = (V1 + V2);

2V = 3V _{1 2.}

Emva iziphumo yokusombulula le nkqubo, siye ndizuze inxaki omnye variable:

1 - 8V = 12V _{1 1;}

V ₁ = _1/20 = _0,05.

Lo kwemveliso kuqala ukusebenza. Ngoku siyakwazi ukufumana ixesha apho ukumelana ngomsebenzi wonke umntu wokuqala:

A = _V-1 * T _1;

1 = 0.05 * T _1;

T ₁ = 20.

Ekubeni ixesha ngeyunithi nganye wamkelwe imini, impendulo ngu: iintsuku ezingama-20.

ngokutsha le ngxaki

Ukuba kakuhle kakhulu izakhono zokusombulula iingxaki kwi intshukumo, kunye neenjongo umsebenzi ufumana ubunzima obuthile, kunokwenzeka ukuba ukusebenza eyokufuna kwezithuthi. Njani? Ukuba uthathe umzekelo lokugqibela, le meko iya kuba ngolu hlobo lulandelayo: Oleg kunye Dima ezihamba omnye komnye, ukuba kwenzeka emva kweeyure ezili-12. Ngokuba bangaphi indlela yokoyisa self Oleg, ukuba uyazi ukuba iiyure ezimbini lipasisa umgama ngendlela elinganayo Dima iiyure ezintathu.