847392299182737373838272728+7382288919198876626272=

Javob: \[ \boxed{847399681471656572714899000} \]

Sonlarni ustun shaklida qo‘shamiz:

\[ \begin{array}{r} 847|392|299|182|737|373|838|272|728\\ 000|007|382|288|919|198|876|626|272\\ \hline 847|399|681|471|656|572|714|899|000 \end{array} \]

Demak:

\[ 847392299182737373838272728+ 7382288919198876626272 = \boxed{847399681471656572714899000}. \]

Suhbatimizdagi avvalgi misol:

\[ 847392299182737373838272728+ 7382288919198876626272 = \boxed{847399681471656572714899000} \]

Mana, maxraji har xil bo‘lgan aralash kasrlarga misollar:

1-misol

\[ 2\frac{1}{3}+1\frac{3}{4} \]

Maxrajlarning EKUKi: \(12\)

\[ \frac13=\frac4{12},\qquad \frac34=\frac9{12} \]

\[ 2\frac4{12}+1\frac9{12} =3\frac{13}{12} = \boxed{4\frac1{12}} \]

2-misol

\[ 5\frac{2}{5}-2\frac{3}{7} \]

Maxrajlarning EKUKi: \(35\)

\[ \frac25=\frac{14}{35},\qquad \frac37=\frac{15}{35} \]

\[ 5\frac{14}{35}-2\frac{15}{35} =\boxed{2\frac{34}{35}} \]

3-misol

\[ 1\frac{5}{6}+2\frac{7}{9} \]

Maxrajlarning EKUKi: \(18\)

\[ \frac56=\frac{15}{18},\qquad \frac79=\frac{14}{18} \]

\[ 1\frac{15}{18}+2\frac{14}{18} =3\frac{29}{18} =\boxed{4\frac{11}{18}} \]

Mana, maxrajlari har xil bo‘lgan aralash kasrlardan 5 ta yangi misol:

1. \[ 3\frac{2}{7}+1\frac{5}{6} =\boxed{5\frac{5}{42}} \]

2. \[ 7\frac{1}{4}-3\frac{2}{9} =\boxed{4\frac{1}{36}} \]

3. \[ 2\frac{3}{8}+4\frac{5}{12} =\boxed{6\frac{19}{24}} \]

4. \[ 9\frac{1}{5}-4\frac{7}{10} =\boxed{4\frac{1}{2}} \]

5. \[ 1\frac{7}{10}+3\frac{11}{15} =\boxed{5\frac{13}{30}} \]

Albatta, mana maxrajlari har xil bo‘lgan aralash kasrlardan 5 ta misol:

1. \[ 3\frac{2}{7}+1\frac{5}{6}=\boxed{5\frac{5}{42}} \]

2. \[ 7\frac{1}{4}-3\frac{2}{9}=\boxed{4\frac{1}{36}} \]

3. \[ 2\frac{3}{8}+4\frac{5}{12}=\boxed{6\frac{19}{24}} \]

4. \[ 9\frac{1}{5}-4\frac{7}{10}=\boxed{4\frac{1}{2}} \]

5. \[ 1\frac{7}{10}+3\frac{11}{15}=\boxed{5\frac{13}{30}} \]

To‘g‘ri yechimi:

\[ 3\frac{2}{7}+1\frac{5}{6} \]

\(7\) va \(6\) ning EKUKi — \(42\).

\[ \frac{2}{7}=\frac{12}{42},\qquad \frac{5}{6}=\frac{35}{42} \]

\[ 3\frac{12}{42}+1\frac{35}{42} =4\frac{47}{42} =4+1\frac{5}{42} =\boxed{5\frac{5}{42}} \]

Demak, \(4\frac{20}{42}\) emas, to‘g‘ri javob:

\[ \boxed{5\frac{5}{42}} \]

Rasmdagi misol:

\[ 2\frac{3}{8}+4\frac{5}{12} \]

1-qadam: Butun qismlarni qo‘shamiz

\[ 2+4=6 \]

2-qadam: Kasrlarning umumiy maxrajini topamiz

Kasrlarning maxrajlari \(8\) va \(12\).

\(8\) va \(12\) ning eng kichik umumiy karralisi:

\[ EKUK(8;12)=24 \]

3-qadam: Kasrlarni umumiy maxrajga keltiramiz

\[ \frac{3}{8}=\frac{3\cdot3}{8\cdot3}=\frac{9}{24} \]

\[ \frac{5}{12}=\frac{5\cdot2}{12\cdot2}=\frac{10}{24} \]

4-qadam: Kasr qismlarini qo‘shamiz

\[ \frac{9}{24}+\frac{10}{24} =\frac{19}{24} \]

5-qadam: Butun va kasr qismlarni birlashtiramiz

\[ 2\frac{3}{8}+4\frac{5}{12} =6+\frac{19}{24} \]

\[ \boxed{6\frac{19}{24}} \]

Javob:

\[ \boxed{6\frac{19}{24}} \]

Albatta, 2- va 3-qadamlarni batafsil tushuntiraman.

Misol:

\[ 2\frac{3}{8}+4\frac{5}{12} \]

2-qadam: Umumiy maxrajni topish

Kasrlarning maxrajlari:

\[ 8 \quad \text{va} \quad 12 \]

Kasrlarni qo‘shish uchun ularning maxrajlari bir xil bo‘lishi kerak. Shuning uchun 8 va 12 ga bo‘linadigan eng kichik sonni topamiz.

8 ning karralari:

\[ 8,\ 16,\ 24,\ 32,\ldots \]

12 ning karralari:

\[ 12,\ 24,\ 36,\ldots \]

Ikkalasida ham uchraydigan eng kichik son — 24.

Demak:

\[ EKUK(8;12)=24 \]

Umumiy maxrajimiz 24 bo‘ladi.

3-qadam: Kasrlarni maxraji 24 ga keltirish

Birinchi kasr:

\[ \frac{3}{8} \]

8 ni 24 ga aylantirish uchun 3 ga ko‘paytiramiz:

\[ 8\cdot3=24 \]

Maxrajni 3 ga ko‘paytirganimiz uchun suratni ham 3 ga ko‘paytiramiz:

\[ \frac{3}{8} = \frac{3\cdot3}{8\cdot3} = \frac{9}{24} \]

Shuning uchun:

\[ \frac{3}{8}=\frac{9}{24} \]

Ikkinchi kasr:

\[ \frac{5}{12} \]

12 ni 24 ga aylantirish uchun 2 ga ko‘paytiramiz:

\[ 12\cdot2=24 \]

Suratni ham 2 ga ko‘paytiramiz:

\[ \frac{5}{12} = \frac{5\cdot2}{12\cdot2} = \frac{10}{24} \]

Shuning uchun:

\[ \frac{5}{12}=\frac{10}{24} \]

Endi ikkala kasrning maxraji bir xil:

\[ \frac{3}{8}+\frac{5}{12} = \frac{9}{24}+\frac{10}{24} \]

Shundan keyin suratlarni qo‘shamiz:

\[ \frac{9+10}{24}=\frac{19}{24} \]

Ya’ni:

\[ 2\frac{3}{8}+4\frac{5}{12} =2+4+\frac{19}{24} =\boxed{6\frac{19}{24}} \]

5 va 10 ga bo‘linadigan eng kichik son — 10.

Chunki:

\[ 10:5=2,\qquad 10:10=1 \]

Demak, EKUK(5; 10)=10.

5 va 10 ga qoldiqsiz bo‘linadigan sonni topamiz.

1. 5 ning karralari:

\[ 5,\ 10,\ 15,\ 20,\ 25,\ 30,\ldots \]

2. 10 ning karralari:

\[ 10,\ 20,\ 30,\ 40,\ldots \]

Ikkala qatorda ham bor sonlar:

\[ 10,\ 20,\ 30,\ldots \]

Ularning ichidagi eng kichik son — 10.

Chunki:

\[ 10:5=2 \]

\[ 10:10=1 \]

Demak:

\[ \boxed{\text{EKUK}(5;10)=10} \]

Kasrlarni umumiy maxrajga keltirganda ham 10 ni olamiz. Masalan:

\[ \frac{3}{5}=\frac{3\cdot2}{5\cdot2}=\frac{6}{10} \]

\[ \frac{7}{10}=\frac{7}{10} \]

Shunday qilib, maxrajlari 5 va 10 bo‘lgan kasrlar uchun umumiy maxraj 10 bo‘ladi.

Agar rasmda \(5\frac{2}{10}\) deb yozgan bo‘lsangiz, bu javob to‘g‘ri emas. 4-misol:

\[ 9\frac15-4\frac7{10} \]

Avval:

\[ \frac15=\frac2{10} \]

Shunda:

\[ 9\frac2{10}-4\frac7{10} \]

\(\frac2{10}<\frac7{10}\) bo‘lgani uchun 9 ta butundan 1 tasini kasrga aylantiramiz:

\[ 9\frac2{10}=8\frac{12}{10} \]

Endi ayiramiz:

\[ 8\frac{12}{10}-4\frac7{10} =4\frac5{10} =4\frac12 \]

Demak, to‘g‘ri javob:

\[ \boxed{4\frac12} \]

4-misolni o‘zbek tilida to‘liq yechamiz:

\[ 9\frac{1}{5}-4\frac{7}{10} \]

1-qadam: Kasrlarning maxrajini tenglashtiramiz

\[ \frac{1}{5}=\frac{2}{10} \]

Shuning uchun:

\[ 9\frac{2}{10}-4\frac{7}{10} \]

2-qadam: Butun sondan qarz olamiz

\(\frac{2}{10}\) dan \(\frac{7}{10}\) ni ayirib bo‘lmaydi. Shuning uchun 9 ta butundan 1 tasini olamiz:

\[ 9\frac{2}{10}=8\frac{12}{10} \]

Chunki:

\[ 1=\frac{10}{10},\qquad \frac{10}{10}+\frac{2}{10}=\frac{12}{10} \]

3-qadam: Ayiramiz

\[ 8\frac{12}{10}-4\frac{7}{10} \]

\[ (8-4)+\left(\frac{12}{10}-\frac{7}{10}\right) =4+\frac{5}{10} \]

\[ \frac{5}{10}=\frac{1}{2} \]

Demak:

\[ \boxed{9\frac15-4\frac7{10}=4\frac12} \]

Javob:

\[ \boxed{4\frac12} \]

Agar 4-misol

\[ 9\frac15-4\frac7{10} \]

bo‘lsa, javob \(5\frac5{10}\) emas, balki:

\[ \frac15=\frac2{10} \]

\[ 9\frac2{10}-4\frac7{10} \]

Kasr qismini ayirish uchun 1 butun qarz olamiz:

\[ 9\frac2{10}=8\frac{12}{10} \]

Endi:

\[ 8\frac{12}{10}-4\frac7{10} =4\frac5{10} \]

Demak, to‘g‘ri javob:

\[ \boxed{4\frac5{10}}=\boxed{4\frac12} \]

\(5\frac5{10}\) bo‘lishi uchun butun qismlar ayirmasi ham 5 bo‘lib qolishi kerak edi, lekin qarz olganimiz sababli \(9\) butun \(8\) ga kamayadi.

Chalkash bo‘lib ketdi, xavotir olmang. Aynan shu misol uchun:

\[ 9\frac15-4\frac7{10} \]

\[ 9\frac15=9\frac2{10} \]

\[ 9\frac2{10}-4\frac7{10} =8\frac{12}{10}-4\frac7{10} =4\frac5{10} \]

Demak, javob:

\[ \boxed{4\frac5{10}}=\boxed{4\frac12} \]

Agar sizdagi 4-misol boshqacha bo‘lsa, uning shartini yoki rasmini yuboring.

Arzimaydi! 😊

Onamning duosi

Bir kuni Anvar maktabdan uyiga keldi. U do‘stlari bilan futbol o‘ynashni juda xohlardi. Shu payt onasi undan:

— Bolam, iltimos, dasturxon yozishga yordam ber, — deb so‘radi.

Anvar avvaliga o‘ynagani chiqmoqchi bo‘ldi. Ammo darsda eshitgan hadisini esladi: Rasululloh sollallohu alayhi vasallamdan “Kimga yaxshi muomala qilishim kerak?” deb so‘ragan odamga uch marta “Onang!” deb javob berilgan, keyin esa “Otang!” deyilgan edi.

Shunda Anvar onasiga yordam berishga qaror qildi. U dasturxon yozdi, non olib keldi va choy damlashda ko‘maklashdi. Onasi xursand bo‘lib:

— Rahmat, bolam. Alloh seni doimo yaxshi farzandlardan qilsin, — deb duo qildi.

Kechqurun otasi ishdan charchab keldi. Anvar otasining kiyimlarini joyiga qo‘ydi va unga suv olib berdi. Otasi o‘g‘lining mehribonligidan mamnun bo‘ldi.

Anvar o‘sha kuni bir narsani chuqur angladi: ota-onaga yaxshi muomala qilish, ularga yordam berish va hurmat ko‘rsatish — har bir farzandning burchidir. Onaning roziligi va otaning duosi inson uchun katta baxtdir.