Question

1. (m-3b) avtomobilning a va b shaharlar orasidagi harakati haqida quyidagilar malum:

• a va b shaharlar orasidagi masofa 276 km;
• avtomobil yoning dastlabki 180 km qismini malum bir tezlikda bosib otdi;
• songra tezligini 12 km/h ga kamaytirib, qolgan masofani shu tezlikda bosib otdi.
• avtomobil a shahardan b shaharga yetib borish uchun jami 5 soat sarfladi.

berilgan malumotlardan foydalanib, avtomobilning dastlabki tezligini toping.
(yechishda 264² = 69696 dan foydalanishingiz mumkin)
a) 72 km/h b) 60 km/h
c) 48 km/h d) 64 km/h

1. (m-3b) tenglamalar sistemasini qanoatlantiruvchi x va y sonlar

juftligi uchun x - y ifoda qiymatini toping: \\(\

$$\begin{cases}x + xy + y = -1\\\\x^2y + xy^2 = -56\\end{cases}$$

\\)
a) \\(-8; 8; 9; -9\\)
b) \\(6; -6; 7; -7\\)
c) \\(6; -6; 9; -9\\)
d) \\(-7; 7; 8; -8\\)

Explanation:

Question 15

Step1: Define variables

Let the initial speed be \( v \) km/h. The distance of the first part is 180 km, so time taken for the first part is \( t_1=\frac{180}{v} \) hours.
The remaining distance is \( 276 - 180=96 \) km. The speed for the second part is \( v + 12 \) km/h, so time taken for the second part is \( t_2=\frac{96}{v + 12} \) hours.

Step2: Total time equation

Total time is 5 hours, so \( t_1 + t_2=5 \). Substituting the expressions for \( t_1 \) and \( t_2 \), we get:

$$ \frac{180}{v}+\frac{96}{v + 12}=5 $$

Multiply through by \( v(v + 12) \) to clear the fractions:

$$ 180(v + 12)+96v = 5v(v + 12) $$

Expand:

$$ 180v+2160 + 96v=5v^{2}+60v $$

Combine like terms:

$$ 276v+2160=5v^{2}+60v $$

Rearrange to form a quadratic equation:

$$ 5v^{2}-216v - 2160 = 0 $$

Divide through by common factor (if any), here we can use quadratic formula \( v=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) where \( a = 5 \), \( b=-216 \), \( c=-2160 \).
First, calculate discriminant \( D=b^{2}-4ac=(-216)^{2}-4\times5\times(-2160)=46656 + 43200=89856 \). Wait, but the problem says we can use \( 264^{2}=69696 \), maybe I made a mistake in calculation. Let's re - check the remaining distance: \( 276-180 = 96 \), correct. Total time is 5 hours. Let's try to factor the quadratic equation again.
From \( \frac{180}{v}+\frac{96}{v + 12}=5 \), multiply both sides by \( v(v + 12) \):
\( 180(v + 12)+96v=5v(v + 12) \)
\( 180v+2160+96v = 5v^{2}+60v \)
\( 276v+2160=5v^{2}+60v \)
\( 5v^{2}-216v - 2160 = 0 \)
Wait, maybe I miscalculated the discriminant. Let's recalculate \( b^{2}-4ac \):
\( b=-216 \), so \( b^{2}=(-216)\times(-216)=46656 \)
\( 4ac = 4\times5\times(-2160)=-43200 \), so \( - 4ac=43200 \)
So \( D = 46656+43200=89856 \). Now, \( \sqrt{89856}=299.76\approx300 \), but this doesn't match the hint. Maybe there is a mistake in the problem - solving approach. Let's try to assume the answer from the options. Let's check option A: \( v = 72 \)
Time for first part: \( \frac{180}{72}=2.5 \) hours. Speed for second part: \( 72 + 12=84 \) km/h. Time for second part: \( \frac{96}{84}=\frac{8}{7}\approx1.14 \) hours. Total time \( 2.5+\frac{8}{7}\approx2.5 + 1.14 = 3.64
eq5 \). Option B: \( v = 60 \)
Time for first part: \( \frac{180}{60}=3 \) hours. Speed for second part: \( 60 + 12 = 72 \) km/h. Time for second part: \( \frac{96}{72}=\frac{4}{3}\approx1.33 \) hours. Total time \( 3+\frac{4}{3}=\frac{13}{3}\approx4.33
eq5 \). Option C: \( v = 48 \)
Time for first part: \( \frac{180}{48}=3.75 \) hours. Speed for second part: \( 48+12 = 60 \) km/h. Time for second part: \( \frac{96}{60}=1.6 \) hours. Total time \( 3.75 + 1.6=5.35
eq5 \). Option D: \( v = 64 \)
Time for first part: \( \frac{180}{64}=\frac{45}{16}\approx2.8125 \) hours. Speed for second part: \( 64 + 12 = 76 \) km/h. Time for second part: \( \frac{96}{76}=\frac{24}{19}\approx1.26 \) hours. Total time \( \approx2.8125+1.26\approx4.07
eq5 \). Wait, maybe I made a mistake in the remaining distance. Wait, \( 276-180 = 96 \), correct. Wait, maybe the speed is decreased? The problem says "tezligini 12 km/h ga kamaytirib" (decrease the speed by 12 km/h). Oh! I misread the problem. It is "tezligini 12 km/h ga kamaytirib" (decrease the speed by 12 km/h), not increase. So speed for the second part is \( v-12 \) km/h. Let's correct that.
So \( t_1=\frac{180}{v} \), \( t_2=\frac{96}{v - 12} \), and \( t_1 + t_2=5 \)
So \( \frac{180}{v}+\frac{96}{v - 12}=5 \)
Multiply through by \( v(v - 12) \):
\( 180(v - 12)+96v=5v(v - 12) \)
Expand:
\( 180v-2160 + 96v=5v^{2}-60v \)
Combine like terms:
\( 276v-2160=5v^{2}-60v \)
R…

Given the system of equations:
\(

$$\begin{cases}x+xy + y=-1\\x^{2}y+xy^{2}=-56\end{cases}$$

\)
Let's factor the second equation: \( xy(x + y)=-56 \)
Let \( u=x + y \) and \( v = xy \)
Then the first equation becomes \( u + v=-1\Rightarrow u=-1 - v \)
The second equation becomes \( v\times u=-56 \)
Substitute \( u=-1 - v \) into \( v\times u=-56 \):
\( v(-1 - v)=-56 \)
\( -v - v^{2}=-56 \)
\( v^{2}+v - 56 = 0 \)
Factor the quadratic equation: \( v^{2}+v - 56=(v + 8)(v - 7)=0 \)
So \( v=-8 \) or \( v = 7 \)
Case 1: If \( v=-8 \), then \( u=-1-(-8)=7 \)
So we have \( x + y=7 \) and \( xy=-8 \)
The quadratic equation is \( t^{2}-7t - 8 = 0 \), factoring: \( (t - 8)(t + 1)=0 \), so \( t = 8 \) or \( t=-1 \)
So \( (x,y)=(8,-1) \) or \( (-1,8) \), then \( x - y=9 \) or \( x - y=-9 \)
Case 2: If \( v = 7 \), then \( u=-1 - 7=-8 \)
So \( x + y=-8 \) and \( xy = 7 \)
The quadratic equation is \( t^{2}+8t + 7 = 0 \), factoring: \( (t + 1)(t + 7)=0 \), so \( t=-1 \) or \( t=-7 \)
So \( (x,y)=(-1,-7) \) or \( (-7,-1) \), then \( x - y = 6 \) or \( x - y=-6 \)
So the possible values of \( x - y \) are \( 6,-6,9,-9 \)

Answer:

B. 60 km/h

Question 16