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22 Umumiy ko’paytuvchini qavsdan tashqariga chiqarish(3 soat)

Author:
October 21st, 2022

7-sinf, algebra.
(3 soatlik bayonnoma)
«_____» -____________, 20___-yil.
Mavzu: Umumiy ko‘paytuvchini qavsdan tashqariga chiqarish.
1. Darsning maqsadi: ko‘phadni birhadga va ko‘phadga ko‘paytirish, birhad va ko‘phadni birhadga
bo‘lish, umumiy ko‘paytuvchini qavsdan tashqariga chiqarish kabi mavzular yuzasidan
tushunchalar hosil qilish, o‘quvchilarning BKM larini shakllantirish, rivojlantirish va
mustahkamlash.
2. Darsning usuli: savol-javob, misol va masalalar yechish.
3. Darsning jihozi: DTS ko‘rgazmalari, tarqatma materiallar, kichik testlar.
DARSNING BORISHI:
1. Tashkiliy qism: o‘quvchilar bilan salomlashish, tozalikni tekshirish, davomatni aniqlash,
o‘quvchilarning dars mashg‘ulotlariga ruhiy jihatda tayyorliklarini aniqlash.
2. O‘tilgan mavzuni so‘rab baholash:
Savol: Ko‘phad nima?
Javob: Birhadning algebraik yig‘indisi ko‘phad deyiladi.
Savol: Ko‘phadning hadlari nima?
Javob: Ko‘phadni tashkil qiluvchi birhadlar ko‘phadning hadlari deyiladi.
Savol: Ko‘phad birhadga qanday ko‘paytiriladi?
Javob: Ko‘phadni birhadga ko‘paytirish uchun ko‘phadning har bir hadini birhadga ko‘paytirish
va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Savol: Ko‘phad ko‘phadga qanday ko‘paytiriladi?
Javob: Ko‘phadni ko‘phadga ko‘paytirish uchun birinchi ko‘phadning har bir hadini ikkinchi
ko‘phadning har bir hadiga ko‘paytirish va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Savol: Ko‘phad birhadga qanday bo‘linadi?
Javob: Ko‘phadni birhadga bo‘lish uchun ko‘phadning har bir hadini shu birhadga bo‘lish va
hosil bo‘lgan natijalarni qo‘shish kerak.
3. O‘tilgan mavzuni mustahkamlasi: o‘quvchilar tushunmagan savollar va ularga tushunarsiz
bo‘lgan jumlalar aniq va hayotiy misollar yordamida tushunturilib beriladi.
4. Yangi mavzuning bayoni:
Qoida: ko‘phadni ikkita yoki bir nechta ko‘phadlar ko‘paytmasi shaklida ifodalash ko‘phadni
ko‘paytuvchilarga ajratish (yoyish) deyiladi.
Masalan: 𝑎2 + 𝑎𝑏 ifodaning 𝑎 = 127 va 𝑏 = 73 bo‘lganda, son qiymatini toping:
Yechilishi: 𝒂𝟐 + 𝒂𝒃 = 𝒂(𝒂 + 𝒃) = 127 ∙ (127 + 73) = 127 ∙ 200 = 25 400.
Qoida: agar ko‘phadning barcha (son yoki harfiy) hadlari umumiy ko‘paytuvchiga ega bo‘lsa, u
holda shu ko‘paytuvchini qavsdan tashqariga chiqarish mumkin. Qavs ichida berilgan ko‘phadni
shu umumiy ko‘paytuvchiga bo‘lish natijasida hosil qilinadigan ko‘phad qoladi.
Masalan: 𝑎𝑏 + 𝑎𝑐 − 𝑎𝑑 ifodaning 𝑎 = 40, 𝑏 = 20, 𝑐 = 15, 𝑑 = 3 bo‘lganda, son qiymatini
toping:
Yechilishi: 𝒂𝒃 + 𝒂𝒄 − 𝒂𝒅 = 𝒂(𝒃 + 𝒄 − 𝒅) = 40 ∙ (20 + 15 − 3) = 40 ∙ 32 = 1 280.
Yodda tuting:
1) Ko‘phadni umumiy ko‘paytuvchini qavsdan tashqariga chiqarish yo‘li bilan ko‘paytuvchilarga
ajratish mumkin. Buning uchun:
– umumiy ko‘paytuvchini topish;
– uni qavsdan tashqariga chiqarish kerak.
Masalan: 21𝑎𝑏 + 42𝑏𝑐 − 7𝑏 ko‘phadni ko‘paytuvchilarga ajrating:
Yechilishi: 21𝑎𝑏 + 42𝑏𝑐 − 7𝑏 = 7𝑏(3𝑎 + 6𝑐 − 1).
2) Umumiy ko‘paytuvchi ko‘phad bo‘lishi ham mumkin.
Masalan: 𝒂(𝒃 + 𝒄) + 𝒏(𝒃 + 𝒄) = (𝒃 + 𝒄)(𝒂 + 𝒏)
Misol va masalalar yechish:
330-misol. Sonlarni tub ko‘paytuvchilarga ajrating:
1) 70 = 2 ∙ 5 ∙ 7;
5) 225 = 32 ∙ 52 .
2) 121 = 11 ∙ 11;
3) 240 = 24 ∙ 3 ∙ 5;
4) 168 = 23 ∙ 3 ∙ 7;
331-misol. Kasrlarni qisqartiring:
1)
45
60
45:15
3
= 60:15 = 4;
18
18:6
3
2) 24 = 24:6 = 4;
3
3)
75∙155
125∙248
3∙5
= 1∙8 =
15
;
8
8
4)
40∙142
17∙153
8∙2
= 1∙3 =
16
.
3
332-misol. Ko‘paytirishning taqsimot qonunini qo‘llang va hisoblang:
1) 81 ∙ 17 − 15 ∙ 81 = 81(17 − 15) = 81 ∙ 2 = 162;
2) 24 ∙ 2,78 + 41 ∙ 2,78 = 2,78(24 + 41) = 2,78 ∙ 65 = 180,7;
3) 15 ∙ 17 + 15 ∙ 67 = 15(17 + 67) = 15 ∙ 84 = 1 260;
3
1
3
1
1
3
3
1
5
5
25
4) 14 8 ∙ 1 4 − 4 8 ∙ 1 4 = 1 4 (14 8 − 4 8) = 1 4 (10 + 0) = 4 ∙ 10 = 2 ∙ 5 = 2 = 12,5.
333-misol. Ko‘paytmani ko‘phad shaklida yozing:
1) (𝑎 + 2)(𝑎 + 3) = 𝑎2 + 3𝑎 + 2𝑎 + 6 = 𝑎2 + 5𝑎 + 6;
3) 3𝑐 3 (2𝑐 3 − 5) = 6𝑐 6 − 15𝑐 3 ;
2) 2𝑥(𝑥 − 1) = 2𝑥 2 − 2𝑥;
4) (𝑎2 + 𝑏)(𝑎 − 𝑏 2 ) = 𝑎3 − 𝑎2 𝑏 2 + 𝑎𝑏 − 𝑏 3 .
334-misol. 𝑨 bekatdan 𝑩 bekatga tomon motorli qayiq 20 𝑘𝑚/𝑠𝑜𝑎𝑡 tezlik bilan jo‘nadi. Oradan
ikki soat o‘tib 𝑨 bekatdan 𝑩 bekatga tomon ikkinchi motorli qayiq 24 𝑘𝑚/𝑠𝑜𝑎𝑡 tezlik bilan
yo‘lga chiqdi. Ikkala qayiq ham 𝑩 bekatga bir vaqtda yetib keldi. 𝑨 va 𝑩 bekatlar orasidagi
masofani toping.
Yechilishi:
𝑆1 = 20 𝑘𝑚/𝑠𝑜𝑎𝑡 ∙ 𝑥 va 𝑆2 = 24 𝑘𝑚/𝑠𝑜𝑎𝑡 ∙ (𝑥 − 2).
Masala shartiga ko‘ra: 𝑆1 = 𝑆2 va 𝑣1 ∙ 𝑡1 = 𝑣2 ∙ 𝑡2 .
Demak, 20 ∙ 𝑥 = 24 ∙ (𝑥 − 2) ga egamiz:
1) 20𝑥 = 24(𝑥 − 2), 20𝑥 = 24𝑥 − 48, 24𝑥 − 20𝑥 = 48, 4𝑥 = 48, 𝑥 = 48: 4, 𝑥 = 12;
2) 20 ∙ 12 = 24 ∙ (12 − 2), 240 = 24 ∙ 10, 240 = 240.
Javob: 𝑆 = 240 𝑘𝑚.
335-misol. 1) 36 + 34 ifodaning 30 ga; 90 ga; 2) 78 + 76 ifodaning 49 ga; 350 ga; 3) 118 − 116
ifodaning 24 ga; 60 ga karrali ekanini isbotlang.
Yechilishi:
1) 36 + 34 = 34 ∙ (32 + 1) = 34 ∙ (9 + 1) = 34 ∙ 10 = 81 ∙ 10 = 810;
810: 30 = 27;
810: 90 = 9.
Javob: Karrali.
2) 78 + 76 = 76 ∙ (72 + 1) = 76 ∙ (49 + 1) = 76 ∙ 50 = 117 649 ∙ 50 = 5 882 450;
5 882 450: 49 = 120 050;
5 882 450: 350 = 16 807.
Javob: Karrali.
3) 118 − 116 = 118−6 = 112 = 121;
1
121: 24 = 5 24;
1
121: 60 = 2 60.
Javob: Karrali emas.
336-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 2𝑚 + 2𝑛 = 2(𝑚 + 𝑛);
4) 6𝑎 + 12 = 6(𝑎 + 2).
2) 3𝑎 − 3𝑥 = 3(𝑎 − 𝑥);
3) 8 − 4𝑥 = 4(2 − 𝑥);
338-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 𝑎𝑥 − 𝑎𝑦 = 𝑎(𝑥 − 𝑦);
4) 3𝑥 − 𝑥𝑦 = 𝑥(3 − 𝑦).
2) 𝑐𝑑 + 𝑏𝑐 = 𝑐(𝑑 + 𝑏);
3) 𝑥𝑦 + 2𝑥 = 𝑥(𝑦 + 2);
339-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 9𝑚𝑛 + 9𝑛 = 9𝑛(𝑚 + 1);
4) 6𝑝𝑘 − 3𝑝 = 3𝑝(2𝑘 − 1).
2) 3𝑏𝑑 − 3𝑎𝑏 = 3𝑏(𝑑 − 𝑎);
3) 11𝑧 − 33𝑦𝑧 = 11𝑧(1 − 3𝑦);
340-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 𝑎𝑏 − 𝑎𝑐 + 𝑎2 = 𝑎(𝑏 − 𝑐 + 𝑎);
3) 6𝑎2 − 3𝑎 + 12𝑏𝑎 = 3𝑎(2𝑎 − 1 + 4𝑏);
2) 𝑥𝑦 − 𝑥 2 + 𝑥𝑧 = 𝑥(𝑦 − 𝑥 + 𝑧);
4) 4𝑏 2 + 8𝑎𝑏 − 12𝑎2 𝑏 = 4𝑏(𝑏 + 2𝑎 − 3𝑎2 ).
341-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 𝑎4 + 2𝑎2 = 𝑎2 (𝑎2 + 2);
2) 𝑎4 − 3𝑎3 = 𝑎3 (𝑎 − 3);
2 3
3 2
2 2 (𝑦
4) 𝑥 𝑦 − 𝑥 𝑦 = 𝑥 𝑦
− 𝑥).
3) 𝑎4 𝑏 2 + 𝑎𝑏 3 = 𝑎𝑏 2 (𝑎3 + 𝑏);
342-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 18𝑦 7 + 12𝑦 4 = 6𝑦 4 (3𝑦 3 + 2);
3) 15𝑥 5 − 5𝑥 3 = 5𝑥 3 (3𝑥 2 − 1);
2) 6𝑥 4 − 24𝑥 2 = 6𝑥 2 (𝑥 2 − 4);
4) 6𝑎5 + 3𝑎2 = 3𝑎2 (2𝑎3 + 1).
343-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 9𝑎2 𝑏 2 − 12𝑎𝑏 3 = 3𝑎𝑏 2 (3𝑎 − 4𝑏);
3) 7𝑎2 𝑏𝑐 + 14𝑎𝑏 2 𝑐 = 7𝑎𝑏𝑐(𝑎 + 2𝑏);
2) 20𝑥 3 𝑦 2 + 4𝑥 2 𝑦 = 4𝑥 2 𝑦(5𝑥𝑦 + 1);
4) 9𝑥𝑦𝑧 2 − 12𝑥𝑦 2 𝑧 = 3𝑥𝑦𝑧(3𝑧 − 4𝑦).
344-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 6𝑦 5 + 12𝑦 4 − 3𝑦 3 = 3𝑦 3 (2𝑦 2 + 4𝑦 − 1); 2) 20𝑎4 − 5𝑎3 + 15𝑎5 = 5𝑎3 (4𝑎 − 1 + 3𝑎2 );
3) 4𝑎2 𝑏 2 + 36𝑎2 𝑏 3 + 6𝑎𝑏 4 = 2𝑎𝑏 2 (2𝑎 + 18𝑎𝑏 + 3𝑏 2 );
4) 2𝑥 2 𝑦 4 − 2𝑥 4 𝑦 2 + 6𝑥 3 𝑦 3 = 2𝑥 2 𝑦 2 (𝑦 2 − 𝑥 2 + 3𝑥𝑦).
346-misol. Ko‘paytuvchilarga ajrating:
1) 𝑎(𝑚 + 𝑛) + 𝑏(𝑚 + 𝑛) = (𝑎 + 𝑏)(𝑚 + 𝑛); 2) 𝑏(𝑎 + 5) − 𝑐(𝑎 + 5) = (𝑏 − 𝑐)(𝑎 + 5);
3) 𝑎(𝑏 − 5) − (𝑏 − 5) = (𝑎 − 1)(𝑏 − 5);
4) (𝑦 − 3) + 𝑏(𝑦 − 3) = (1 + 𝑏)(𝑦 − 3).
347-misol. Ko‘paytuvchilarga ajrating:
1) 2𝑎(𝑎 − 𝑏) + 3𝑏(𝑎 − 𝑏) = (2𝑎 + 3𝑏)(𝑎 − 𝑏);
3) 5𝑎(𝑥 + 𝑦) − 4𝑏(𝑥 + 𝑦) = (5𝑎 − 4𝑏)(𝑥 + 𝑦);
2) 3𝑛(𝑚 − 3) + 5𝑚(𝑚 − 3) = (3𝑛 + 5𝑚)(𝑚 − 3);
4) 7𝑎(𝑐 − 𝑑) − 2𝑏(𝑐 − 𝑑) = (7𝑎 − 2𝑏)(𝑐 − 𝑑).
348-misol. Ko‘paytuvchilarga ajrating:
1) 𝑎2 (𝑥 − 𝑦) + 𝑏 2 (𝑥 − 𝑦) = (𝑎2 + 𝑏 2 )(𝑥 − 𝑦); 2) 𝑎2 (𝑥 + 𝑦) − 𝑏 2 (𝑥 + 𝑦) = (𝑎2 − 𝑏 2 )(𝑥 + 𝑦);
3) 𝑎(𝑥 2 + 𝑦 2 ) − 𝑏(𝑥 2 + 𝑦 2 ) = (𝑎 − 𝑏)(𝑥 2 + 𝑦 2 ); 4) 𝑥(𝑎2 − 2𝑏 2 ) + 𝑦(𝑎2 − 2𝑏 2 ) = (𝑥 + 𝑦)(𝑎2 − 2𝑏 2 ).
349-misol. Ko‘paytuvchilarga ajrating:
1) 2𝑏(𝑥 − 1) − 3𝑎(𝑥 − 1) + 𝑐(𝑥 − 1) = (2𝑏 − 3𝑎 + 𝑐)(𝑥 − 1);
2) 𝑐(𝑝 − 𝑞) − 𝑎(𝑝 − 𝑞) + 𝑑(𝑝 − 𝑞) = (𝑐 − 𝑎 + 𝑑)(𝑝 − 𝑞);
3) 𝑥(𝑎2 + 𝑏 2 ) + 𝑦(𝑎2 + 𝑏 2 ) − 𝑧(𝑎2 + 𝑏 2 ) = (𝑥 + 𝑦 − 𝑧)(𝑎2 + 𝑏 2 );
4) 𝑚(𝑥 2 + 1) − 𝑛(𝑥 2 + 1) − 𝑙(𝑥 2 + 1) = (𝑚 − 𝑛 − 𝑙)(𝑥 2 + 1).
350-misol. Ko‘paytuvchilarga ajrating:
1) 𝑐(𝑎 − 𝑏) + 𝑏(𝑏 − 𝑎) = 𝑐(𝑎 − 𝑏) − 𝑏(𝑎 − 𝑏) = (𝑐 − 𝑏)(𝑎 − 𝑏);
2) 𝑎(𝑏 − 𝑐) − 𝑐(𝑐 − 𝑏) = 𝑎(𝑏 − 𝑐) + 𝑐(𝑏 − 𝑐) = (𝑎 + 𝑐)(𝑏 − 𝑐);
3) (𝑥 − 𝑦) + 𝑏(𝑦 − 𝑥) = (𝑥 − 𝑦) − 𝑏(𝑥 − 𝑦) = (1 − 𝑏)(𝑥 − 𝑦);
4) 2𝑏(𝑥 − 𝑦) − (𝑦 − 𝑥) = 2𝑏(𝑥 − 𝑦) + (𝑥 − 𝑦) = (2𝑏 + 1)(𝑥 − 𝑦).
351-misol. Ko‘paytuvchilarga ajrating:
1) 7(𝑦 − 3) − 𝑎(3 − 𝑦) = 7(𝑦 − 3) + 𝑎(𝑦 − 3) = (7 + 𝑎)(𝑦 − 3);
2) 6(𝑎 − 2) + 𝑎(2 − 𝑎) = 6(𝑎 − 2) − 𝑎(𝑎 − 2) = (6 − 𝑎)(𝑎 − 2);
3) 𝑏 2 (𝑎 − 1) − 𝑐(1 − 𝑎) = 𝑏 2 (𝑎 − 1) + 𝑐(𝑎 − 1) = (𝑏 2 + 𝑐)(𝑎 − 1);
4) 𝑎2 (𝑚 − 2) + 𝑏(2 − 𝑚) = 𝑎2 (𝑚 − 2) − 𝑏(𝑚 − 2) = (𝑎2 − 𝑏)(𝑚 − 2).
352-misol. Ko‘paytuvchilarga ajrating:
1) 𝑎(𝑏 − 𝑐) + 𝑏 2 (𝑏 − 𝑐) − 7(𝑐 − 𝑏) = 𝑎(𝑏 − 𝑐) + 𝑏 2 (𝑏 − 𝑐) + 7(𝑏 − 𝑐) = (𝑎 + 𝑏 2 + 7)(𝑏 − 𝑐);
2) 𝑥(𝑥 − 𝑦) + 𝑦(𝑦 − 𝑥) − 3(𝑥 − 𝑦) = 𝑥(𝑥 − 𝑦) − 𝑦(𝑥 − 𝑦) − 3(𝑥 − 𝑦) = (𝑥 − 𝑦 − 3)(𝑥 − 𝑦);
3) 𝑥(𝑎 − 2) + 𝑦(2 − 𝑎) + (2 − 𝑎) = 𝑥(𝑎 − 2) − 𝑦(𝑎 − 2) − (𝑎 − 2) = (𝑥 − 𝑦 − 1)(𝑎 − 2);
4) 𝑎(𝑏 − 3) + (3 − 𝑏) − 𝑏(3 − 𝑏) = −𝑎(3 − 𝑏) + (3 − 𝑏) − 𝑏(3 − 𝑏) = (−𝑎 + 1 − 𝑏)(3 − 𝑏).
5. O‘tilgan mavzuni mustahkamlash: o‘quvchilar tushunmagan savollarni aniq misollar yordamida
tushuntiraman.
6. O‘tilgan mavzuni so‘rab baholash:
Savol: Ko‘phad nima?
Javob: Birhadning algebraik yig‘indisi ko‘phad deyiladi.
Savol: Ko‘phadning hadlari nima?
Javob: Ko‘phadni tashkil qiluvchi birhadlar ko‘phadning hadlari deyiladi.
Savol: Ko‘phad birhadga qanday ko‘paytiriladi?
Javob: Ko‘phadni birhadga ko‘paytirish uchun ko‘phadning har bir hadini birhadga
ko‘paytirish va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Savol: Ko‘phad ko‘phadga qanday ko‘paytiriladi?
Javob: Ko‘phadni ko‘phadga ko‘paytirish uchun birinchi ko‘phadning har bir hadini ikkinchi
ko‘phadning har bir hadiga ko‘paytirish va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Savol: Ko‘phad birhadga qanday bo‘linadi?
Javob: Ko‘phadni birhadga bo‘lish uchun ko‘phadning har bir hadini shu birhadga bo‘lish va
hosil bo‘lgan natijalarni qo‘shish kerak.
Savol: Ko‘phadni ko‘paytuvchilarga ajratish deb nimaga aytiladi?
Javob: Ko‘phadni ikkita yoki bir nechta ko‘phadlar ko‘paytmasi shaklida ifodalash ko‘phadni
ko‘paytuvchilarga ajratish (yoyish) deyiladi.
7. Uyga vazifa: O‘tilgan mavzuni o‘qib o‘rganish va misollar yechish.
337-misol. Umumiy ko‘paytuvchini qavsdan tashqariga chiqaring:
1) 9𝑎 + 12𝑏 + 3 = 3(3𝑎 + 4𝑏 + 1);
3) − 10𝑥 + 15𝑦 − 5𝑧 = −5(2𝑥 − 3𝑦 + 𝑧);
2) 8𝑎 − 4𝑏 − 2 = 2(4𝑎 − 2𝑏 − 1);
4) 9𝑥 − 3𝑦 + 12𝑧 = 3(3𝑥 − 𝑦 + 4𝑧).
345-misol. Hisoblang:
1) 1372 + 137 ∙ 63 = 137(137 + 63) = 137 ∙ 200 = 27 400;
2) 1872 − 187 ∙ 87 = 187(187 − 87) = 187 ∙ 100 = 18 700;
3) 0,73 + 0,7 ∙ 9,51 = 0,7(0,72 + 9,51) = 0,7(0,49 + 9,51) = 0,7 ∙ 10 = 7;
4) 0,93 − 0,81 ∙ 2,9 = 0,81(0,9 − 2,9) = 0,81 ∙ (−2) = −1,62.
353-misol. Tenglamani yeching:
1) 8 − (𝑥 − 3)(𝑥 + 3) = 10 − (𝑥 − 1)2
8 − (𝑥 2 − 9) = 10 − (𝑥 2 − 2𝑥 + 1)
8 − 𝑥 2 + 9 = 10 − 𝑥 2 + 2𝑥 − 1
−𝑥 2 + 𝑥 2 − 2𝑥 = 10 − 1 − 8 − 9
−2𝑥 = −8 /: (−2)
𝑥 = 4;
1
3) 𝑥: 15 = 2 12 : 14,5
25 145
10
𝑥: 15 = 12 :
5
𝑥: 15 =
25
∙
105
612 14529
2) (2𝑥 + 1)2 − (2𝑥 − 3)2 = 4(7𝑥 − 5)
4𝑥 2 + 4𝑥 + 1 − (4𝑥 2 − 12𝑥 + 9) = 28𝑥 − 20
4𝑥 2 + 4𝑥 + 1 − 4𝑥 2 + 12𝑥 − 9 = 28𝑥 − 20
4𝑥 2 + 4𝑥 − 4𝑥 2 + 12𝑥 − 28𝑥 = −20 + 9 − 1
−12𝑥 = −12 /: (−12)
𝑥 = 1;
4)
𝑥
2,3
=
23
2,1
9
6
7
21 69
7
𝑥: 10 = 10 :
𝑥:
23
10
7
=
21
10
∙
7
6923
5∙5
𝑥: 15 = 6∙29
𝑥: 15 =
𝑥=
𝑥=
𝑥=
𝑥=
25
174
25
58174
25∙5
58
125
58
9
2 58
∙ 155
23
7
7
𝑥: 10 = 10 ∙ 23
𝑥:
23
10
𝑥=
=
49
230
49
10230
49∙1
𝑥 = 10∙10
49
𝑥 = 100
𝑥 = 0,49.
∙
231
10
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