Question

(b-tilish; qoglash; m-mulohaza) oid test savollari

1. (b-2b) ((x - 5)^2 = 0) tenglamaga teng kuchli bolmagan tenglamani aniqlang.

a) (\frac{x^2 - 25}{x + 5} = 0)
b) (ysqrt{x - 1} = 2)
c) (3x - 15 = 0)
d) (2x + 10 = 0)

1. (b-2b) (x = 1), (y = 3) sonlar quyidagi tenglamalar sistemining qaysi birining yechimi bola oladi?

a) (\begin{cases} 3xy = 9 \\ 2y + x = 5 end{cases})
b) (\begin{cases} 2xy = 6 \\ 2y - x = 7 end{cases})
c) (\begin{cases} 3xy = 10 \\ x - 2y = -5 end{cases})
d) (\begin{cases} 2xy = 6 \\ x + 3y = 10 end{cases})

Explanation:

Question 1

Step 1: Analyze the original equation

The original equation is \((x - 5)^2=0\), which simplifies to \(x - 5=0\) (since a square of a number is zero only when the number itself is zero), so \(x = 5\). Now we check each option:

• Option A: \(\frac{x^2 - 25}{x + 5}=0\). First, factor the numerator: \(x^2-25=(x - 5)(x + 5)\). So the equation becomes \(\frac{(x - 5)(x + 5)}{x + 5}=0\). But we must note that \(x

eq - 5\) (since the denominator cannot be zero). Canceling out \(x + 5\) (for \(x
eq - 5\)), we get \(x-5 = 0\), so \(x = 5\). But the domain of this function excludes \(x=-5\), while the original equation \((x - 5)^2 = 0\) has \(x = 5\) as a solution (with multiplicity two) and no restriction on \(x=-5\) (since \((x - 5)^2\) is defined for all \(x\)). Wait, maybe I made a mistake. Wait, the original equation \((x - 5)^2=0\) is a quadratic equation with a repeated root at \(x = 5\). Let's check the other options:

• Option C: \(3x-15 = 0\). Solve for \(x\): \(3x=15\), so \(x = 5\). This is a linear equation with solution \(x = 5\), same as the original equation's solution. But wait, the original equation is a quadratic, and option C is linear. Wait, maybe the question is about equivalent equations (same solution set). Let's check the solution sets:

Original equation: \((x - 5)^2=0\) has solution \(x = 5\) (multiplicity 2, but as a set, \(\{5\}\)).

Option A: \(\frac{x^2 - 25}{x + 5}=0\). The numerator is zero when \(x^2-25 = 0\) i.e., \(x=\pm5\), but denominator is zero when \(x=-5\), so the solution is \(x = 5\) (since \(x=-5\) is excluded). So the solution set is \(\{5\}\), same as original? Wait, but the original equation is defined for \(x=-5\) (since \((-5 - 5)^2=(-10)^2 = 100
eq0\)), but in option A, \(x=-5\) is not in the domain. So the solution sets are the same (both have \(x = 5\) as solution), but the domains are different? Wait, maybe the question is about equivalent equations in terms of solutions (ignoring domain restrictions for rational equations? Maybe in the context of the problem, they consider the solution \(x = 5\) only). Let's check option C: \(3x-15=0\) simplifies to \(x = 5\), same solution. Option D: \(2x+10=0\) gives \(x=-5\), which is not a solution of the original equation. Option B: \(\sqrt{x - 1}=2\) gives \(x - 1 = 4\) so \(x = 5\), but this is a square root equation, and the original is a quadratic. Wait, maybe the correct answer is C? Wait, let's re - evaluate:

The original equation \((x - 5)^2=0\) can be expanded as \(x^2-10x + 25=0\). Option C: \(3x-15 = 0\) is \(x = 5\), which is the solution of the original equation. Option A: \(\frac{x^2 - 25}{x + 5}\) is equal to \(x - 5\) when \(x
eq - 5\), so the equation \(\frac{x^2 - 25}{x + 5}=0\) is equivalent to \(x - 5=0\) with \(x
eq - 5\), while the original equation has \(x = 5\) as a solution (and \(x=-5\) is not a solution of the original equation, since \((-5 - 5)^2=100
eq0\)). So the solution set of option A is \(\{5\}\) (since \(x=-5\) is excluded, and \(x = 5\) is the solution), and the original equation's solution set is \(\{5\}\) (since \((5 - 5)^2=0\) and \((-5 - 5)^2=100
eq0\)). Wait, maybe the problem is about having the same solution. Let's check the linear equations:

Option C: \(3x-15=0\) \(\Rightarrow\) \(x = 5\), same as the solution of \((x - 5)^2=0\) (which is \(x = 5\)). Option A: The rational equation has the same solution \(x = 5\) but with a restricted domain, but maybe in the context of the problem, they consider the solution \(x = 5\) only. Wait, maybe the correct answer is C? Wait, let's check the other options…

Step 1: Check option A

For system A: \(

$$\begin{cases}3xy=9\\2y + x=5\end{cases}$$

\)

• First equation: \(3xy=3\times1\times3 = 9\), which satisfies the first equation.
• Second equation: \(2y+x=2\times3 + 1=6 + 1=7

eq5\). So option A is not correct.

Step 2: Check option B

For system B: \(

$$\begin{cases}2xy=6\\2y - x=7\end{cases}$$

\)

• First equation: \(2xy=2\times1\times3 = 6\), which satisfies the first equation.
• Second equation: \(2y - x=2\times3-1 = 6 - 1=5

eq7\). So option B is not correct.

Step 3: Check option C

For system C: \(

$$\begin{cases}3xy=10\\x - 2y=-5\end{cases}$$

\)

• First equation: \(3xy=3\times1\times3 = 9

eq10\). So option C is not correct.

Step 4: Check option D

For system D: \(

$$\begin{cases}2xy=6\\x + 3y=10\end{cases}$$

\)

• First equation: \(2xy=2\times1\times3 = 6\), which satisfies the first equation.
• Second equation: \(x + 3y=1+3\times3=1 + 9 = 10\), which satisfies the second equation.

Answer:

C

Question 2

We are given \(x = 1\) and \(y = 3\), and we need to check which system of equations is satisfied by these values.