Реши задание на изображении
Qisqa javob:
- A 2) A 3) A 4) A 5) A 6) B
1)
\[
x^2-6x+8=(x-2)(x-4)
\]
Maxraj nolga teng bo‘lmasligi kerak:
\[
x\ne 2,\quad x\ne 4
\]
Demak:
\[
(-\infty;2)\cup(2;4)\cup(4;\infty)
\]
Javob: A.
2)
Surat:
\[
8a^3-36a^2+54a-27=(2a-3)^3
\]
Maxraj:
\[
4a^2-12a+9=(2a-3)^2
\]
Qisqartiramiz:
\[
\frac{(2a-3)^3}{(2a-3)^2}=2a-3
\]
Javob: A. \((a\ne \frac32)\)
3)
Belgilaymiz:
\[
x=a+b,\quad y=b+c,\quad z=a+c
\]
Unda:
\[
x+y+z=2(a+b+c)=10
\]
Berilgan:
\[
\frac1x+\frac1y+\frac1z=\frac{13}{5}
\]
Masalan, \(c=5-x\), shuning uchun:
\[
\frac{c}{a+b}=\frac{5-x}{x}=\frac5x-1
\]
Xuddi shunday qolgan ikkitasi ham. Demak:
\[
\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}
=5\left(\frac1x+\frac1y+\frac1z\right)-3
\]
\[
=5\cdot\frac{13}{5}-3=13-3=10
\]
Javob: A.
4)
Bo‘lish bajaramiz:
\[
\frac{n^2+3n-15}{n-3}=n+6+\frac{3}{n-3}
\]
Butun natural son bo‘lishi uchun \(n-3\) soni \(3\) ni bo‘lishi kerak:
\[
n-3=\pm1,\ \pm3
\]
Natural \(n\) lar:
\[
n=2,\ 4,\ 6
\]
Yig‘indi:
\[
2+4+6=12
\]
Javob: A.
5)
Avval qavs ichidagi ifodani soddalashtiramiz:
\[
\frac1{(x-1)^2}-\frac{x}{1-x^2}
=\frac1{(1-x)^2}-\frac{x}{(1-x)(1+x)}
\]
Umumiy maxraj:
\[
=\frac{1+x-x+x^2}{(1-x)^2(1+x)}
=\frac{1+x^2}{(1-x)^2(1+x)}
\]
Endi ko‘paytiramiz:
\[
\frac{1-x^2}{1+x^2}\cdot \frac{1+x^2}{(1-x)^2(1+x)}
=\frac{1}{1-x}
\]
Natijada:
\[
\frac1{1-x}-\frac1{1-x}=0
\]
Javob: A. \((x\ne \pm1)\)
6)
Asosiy qoida:
\[
\frac{a-1}{b+1}=\frac{1-a}{-b-1}
\]
ya’ni surat va maxrajni bir vaqtda \(-1\) ga ko‘paytirish mumkin. Faqat bittasining ishorasini o‘zgartirish esa ifodani o‘zgartiradi.
Tekshiruv:
- noto‘g‘ri
- noto‘g‘ri
- to‘g‘ri
- to‘g‘ri
Demak: 1-noto‘g‘ri, 2-noto‘g‘ri, 3-to‘g‘ri, 4-to‘g‘ri
Javob: B.
Обсуждение
3 сообщенияIshlanishi va tushuntir
Javoblar:
1) A 2) A 3) A 4) A 5) A 6) B
Quyida har bir masalaning ishlanishi va tushuntirishi.
1) Funksiyaning aniqlanish sohasi
Maxraj nolga teng bo‘lmasligi kerak:
\[ x^2-6x+8\ne 0 \]
Kvadrat uchhadni ko‘paytuvchilarga ajratamiz:
\[ x^2-6x+8=(x-2)(x-4) \]
Demak:
\[ (x-2)(x-4)\ne0 \]
Bundan:
\[ x\ne2,\qquad x\ne4 \]
Shuning uchun aniqlanish sohasi:
\[ (-\infty;2)\cup(2;4)\cup(4;+\infty) \]
Javob: A.
2) Kasrni qisqartirish
Surat:
\[ 8a^3-36a^2+54a-27 \]
Bu ifoda to‘liq kub formulasi:
\[ (2a-3)^3=8a^3-36a^2+54a-27 \]
Maxraj esa:
\[ 4a^2-12a+9=(2a-3)^2 \]
Unda:
\[ \frac{(2a-3)^3}{(2a-3)^2}=2a-3 \]
Bunda \(2a-3\ne0\), ya’ni:
\[ a\ne\frac32 \]
Javob: A.
3) Belgilash usuli bilan yechish
Berilgan:
\[ a+b+c=5 \]
va
\[ \frac1{a+b}+\frac1{b+c}+\frac1{a+c}=\frac{13}{5} \]
Belgilaymiz:
\[ x=a+b,\qquad y=b+c,\qquad z=a+c \]
Unda:
\[ x+y+z=2(a+b+c)=2\cdot5=10 \]
Endi kerakli ifodani ko‘ramiz:
\[ \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c} \]
\(c=5-(a+b)=5-x\), shuning uchun:
\[ \frac{c}{a+b}=\frac{5-x}{x}=\frac5x-1 \]
Xuddi shunday:
\[ \frac{a}{b+c}=\frac5y-1,\qquad \frac{b}{a+c}=\frac5z-1 \]
Ularni qo‘shamiz:
\[ \left(\frac5x+\frac5y+\frac5z\right)-3 \]
\[ =5\left(\frac1x+\frac1y+\frac1z\right)-3 \]
Berilganiga ko‘ra:
\[ \frac1x+\frac1y+\frac1z=\frac{13}{5} \]
Demak:
\[ 5\cdot\frac{13}{5}-3=13-3=10 \]
Javob: A.
4) Butun son shartidan foydalanish
Ifodani bo‘lamiz:
\[ \frac{n^2+3n-15}{n-3} \]
Bo‘lishni bajaramiz:
\[ n^2+3n-15=(n-3)(n+6)+3 \]
Shuning uchun:
\[ \frac{n^2+3n-15}{n-3} =n+6+\frac{3}{n-3} \]
Bu ifoda butun son bo‘lishi uchun:
\[ \frac{3}{n-3} \]
butun son bo‘lishi kerak. Demak \(n-3\) soni \(3\) ni bo‘lishi kerak:
\[ n-3=\pm1,\ \pm3 \]
Bundan:
\[ n=4,\ 2,\ 6,\ 0 \]
Natural son sifatida odatda \(0\) olinmaydi, shuning uchun:
\[ n=2,\ 4,\ 6 \]
Yig‘indi:
\[ 2+4+6=12 \]
Javob: A.
Eslatma: ba’zi darsliklarda \(0\) ham natural son deyilsa, u qo‘shilganda ham yig‘indi o‘zgarmaydi: \(0+2+4+6=12\).
5) Ifodani soddalashtirish
Berilgan ifoda:
\[ \left(\frac1{(x-1)^2}-\frac{x}{1-x^2}\right)\cdot \frac{1-x^2}{1+x^2}-\frac1{1-x} \]
Avval qavs ichini soddalashtiramiz:
\[ \frac1{(x-1)^2}=\frac1{(1-x)^2} \]
va
\[ 1-x^2=(1-x)(1+x) \]
Unda:
\[ \frac1{(1-x)^2}-\frac{x}{(1-x)(1+x)} \]
Umumiy maxraj:
\[ (1-x)^2(1+x) \]
Birinchi kasr:
\[ \frac{1+x}{(1-x)^2(1+x)} \]
Ikkinchi kasr:
\[ \frac{x(1-x)}{(1-x)^2(1+x)} \]
Ayirish natijasi:
\[ \frac{1+x-x+x^2}{(1-x)^2(1+x)} = \frac{1+x^2}{(1-x)^2(1+x)} \]
Endi bu ifodani \(\frac{1-x^2}{1+x^2}\) ga ko‘paytiramiz:
\[ \frac{1+x^2}{(1-x)^2(1+x)} \cdot \frac{(1-x)(1+x)}{1+x^2} \]
Qisqartirishdan keyin:
\[ \frac1{1-x} \]
Demak butun ifoda:
\[ \frac1{1-x}-\frac1{1-x}=0 \]
Bunda \(x\ne1\), \(x\ne-1\), chunki maxrajlar nolga aylanadi.
Javob: A. Ya’ni natija \(0\).
6) Kasr ishorasi qoidasi
Asosiy qoida:
\[ \frac{A}{B}=\frac{-A}{-B} \]
Ya’ni surat va maxrajning ishorasini birgalikda o‘zgartirish kasr qiymatini o‘zgartirmaydi.
Agar faqat surat yoki faqat maxraj ishorasi o‘zgartirilsa, kasr ishorasi o‘zgaradi:
\[ \frac{-A}{B}=-\frac AB \]
\[ \frac A{-B}=-\frac AB \]
Shuning uchun:
- suratning ishorasini alohida o‘zgartirish — noto‘g‘ri;
- maxrajning ishorasini alohida o‘zgartirish — noto‘g‘ri;
- surat va maxraj ishorasini birga o‘zgartirish — to‘g‘ri;
- shu qoidaga mos yozilgan tenglik — to‘g‘ri.
Demak oldingi yechimdagi tartibda:
\[ 1-\text{noto‘g‘ri},\quad 2-\text{noto‘g‘ri},\quad 3-\text{to‘g‘ri},\quad 4-\text{to‘g‘ri} \]
Javob: B.
Yakuniy javoblar:
1-A, 2-A, 3-A, 4-A, 5-A, 6-B.
