Boshlang`

Yechish: Ko`rinib turibdiki, bu yerda dastlabki 50 ta natural sonlar ketma-ket yozilib, ko`p xonali bitta son hosil qilingan. 1 dan 50 gacha bo`lgan barcha natural sonlarning raqamlari yig`indisi berilgan sonning raqamlari yig`indisiga teng.
_{1 dan 10 gacha: 1+2+3+4+5+6+7+8+9+1+0=46; ––bu yerda 46 kalit vazifasini bajaradi}

.	11 dan 20 gacha: 56;
.	21 dan 30 gacha: 66;	+
.	31 dan 40 gacha: 76;
.	41 dan 50 gacha: 86
.	____		Javob: 330. (B)
.	330

Topshiriq: Ushbu 31323334…7980 sonning raqamlari yig`indisini toping.
A) 473 B) 480 C) 460 D) 490 E) 453 Javob: 480. (B)
6-misol. 4<sup>12</sup>+4<sup>12</sup>+4<sup>12</sup>+4<sup>12</sup> yig`indining yarmini hisoblang.
_{A) 2}²⁵ _{B) 2}²⁴ _{C) 4}⁴⁸ _{D) 2 ∙ 4}¹⁶ _{E) 4}²⁵
_Yechish:

Topshiriq: 4¹³+4¹³+4¹³+4¹³ yig`indining yarmini hisoblang.
A) 224 B) 225 C) 8 ∙ 412 D) 448
<sup>7-misol. 4–7+8–11+12–15+…+96–99 yig`indini hisoblang.</sup>
A) –75 B) –80 C) –72 D) –63
Yechish: 4–7+8–11+12–15+…+96–99=(4+8+12+…+92+96)–(7+11+15+…+95+99) =>
=> Birinchi qavsdagi qo`shiluvchilar soni: 96:4=24 ta, demak, ikkinchi qavsda ham 24 ta
<sup>=> ((4+96)+(8+92)+…)–((7+99)+(11+95)+…)=100∙12–106∙12=12∙(100–106)=12∙(–6)=
= –72. Javob: –72. (C)</sup>
Topshiriq: Yig`indini hisoblang: 1–2+3–4+5–6+…+993.
A) 497 B) 496 C) –497 D) –496. Javob: 497. (A)
BOSHLANG`ICH TUSHUNCHALAR
1. Matematika so`zining ma`nosi nima?
2. Matematikaning alifbosi nima?
3. Sanoqda foydalaniladigan sonlar nima deb ataladi?


^{4. 5– raqammi yoki sonmi?}
5. 17–raqammi yoki sonmi?


6. Juft sonlarni ifodalovchi forlmulani yozing.
7. 2019 sonini toq son formulasida tasvirlang.
8. Tub sonlardan nechtasi juft son?
9. 100 gacha nechta tub son bor?
10. Tub sonlar ko`paytmasi tubmi yoki murakkabmi?
11. Mukammal son formulasini yozing.
12. To`rtinchi mukammal sonni aniqlang.
13. Ko`paytirishning qo`shishga nisbatan taqsimot qonuni yozing.

1996-2007 Matematika 1-bob.1-mavzu M.H.Sharopov.