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Sagot :
To determine if the proportion \(\frac{v}{x} = \frac{x}{v+w}\) is true, we'll start by cross-multiplying to check if both sides are equal.
[tex]\[ \frac{v}{x} = \frac{x}{v+w} \][/tex]
Cross-multiplication gives us:
[tex]\[ v \cdot (v + w) = x \cdot x \][/tex]
Simplifying both sides:
[tex]\[ v^2 + vw = x^2 \][/tex]
For this proportion \(\frac{v}{x} = \frac{x}{v+w}\) to be true, the equation \(v^2 + vw = x^2\) must hold.
Since we do not have specific values for \(v\), \(w\), and \(x\), we cannot definitively say this proportion is true in the general sense without more information. In general, it is unlikely for \(v^2 + vw\) to equal \(x^2\) for arbitrary values of \(v\), \(w\), and \(x\).
Therefore, the given proportion is generally false. Let's check the corrections provided in options B and C.
Option B:
[tex]\[ \frac{w}{x} = \frac{x}{v+w} \][/tex]
Cross-multiplying gives us:
[tex]\[ w \cdot (v + w) = x \cdot x \][/tex]
Simplifying both sides:
[tex]\[ wv + w^2 = x^2 \][/tex]
Again, without specific values, we cannot generalize this as true. It has a similar issue to the original proportion.
Option C:
[tex]\[ \frac{\nu}{\nu} = \frac{y}{\nu+w} \][/tex]
Since \(\frac{\nu}{\nu} = 1\) for any non-zero value of \(\nu\), this simplifies to:
[tex]\[ 1 = \frac{y}{\nu+w} \][/tex]
Cross-multiplying or simplifying:
[tex]\[ \nu + w = y \][/tex]
So, \(y = \nu + w\). This holds true only if \(y\) is explicitly defined as the sum of \(\nu\) and \(w\). Without a given context or specific values, one cannot assume this is generally true either.
Since none of the proportions hold true for arbitrary values, the answer to whether the given proportion \(\frac{v}{x}=\frac{x}{v+w}\) is true is:
B. false (And with no correction indicating a universally correct proportion based on given options).
[tex]\[ \frac{v}{x} = \frac{x}{v+w} \][/tex]
Cross-multiplication gives us:
[tex]\[ v \cdot (v + w) = x \cdot x \][/tex]
Simplifying both sides:
[tex]\[ v^2 + vw = x^2 \][/tex]
For this proportion \(\frac{v}{x} = \frac{x}{v+w}\) to be true, the equation \(v^2 + vw = x^2\) must hold.
Since we do not have specific values for \(v\), \(w\), and \(x\), we cannot definitively say this proportion is true in the general sense without more information. In general, it is unlikely for \(v^2 + vw\) to equal \(x^2\) for arbitrary values of \(v\), \(w\), and \(x\).
Therefore, the given proportion is generally false. Let's check the corrections provided in options B and C.
Option B:
[tex]\[ \frac{w}{x} = \frac{x}{v+w} \][/tex]
Cross-multiplying gives us:
[tex]\[ w \cdot (v + w) = x \cdot x \][/tex]
Simplifying both sides:
[tex]\[ wv + w^2 = x^2 \][/tex]
Again, without specific values, we cannot generalize this as true. It has a similar issue to the original proportion.
Option C:
[tex]\[ \frac{\nu}{\nu} = \frac{y}{\nu+w} \][/tex]
Since \(\frac{\nu}{\nu} = 1\) for any non-zero value of \(\nu\), this simplifies to:
[tex]\[ 1 = \frac{y}{\nu+w} \][/tex]
Cross-multiplying or simplifying:
[tex]\[ \nu + w = y \][/tex]
So, \(y = \nu + w\). This holds true only if \(y\) is explicitly defined as the sum of \(\nu\) and \(w\). Without a given context or specific values, one cannot assume this is generally true either.
Since none of the proportions hold true for arbitrary values, the answer to whether the given proportion \(\frac{v}{x}=\frac{x}{v+w}\) is true is:
B. false (And with no correction indicating a universally correct proportion based on given options).
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