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Sagot :
To solve the equation
[tex]\[ \frac{k}{k+5} + \frac{5}{7-k} = 1 \][/tex]
we will walk through the process in a structured manner.
1. Combine the fractions on the left-hand side:
To do this, we need a common denominator. The common denominator for the fractions [tex]\(\frac{k}{k+5}\)[/tex] and [tex]\(\frac{5}{7-k}\)[/tex] is [tex]\((k+5)(7-k)\)[/tex].
[tex]\[ \frac{k}{k+5} + \frac{5}{7-k} = \frac{k(7-k) + 5(k+5)}{(k+5)(7-k)} \][/tex]
2. Expand and simplify the numerator:
[tex]\[ k(7-k) + 5(k+5) = 7k - k^2 + 5k + 25 = 12k - k^2 + 25 \][/tex]
So, the equation now looks like this:
[tex]\[ \frac{12k - k^2 + 25}{(k+5)(7-k)} = 1 \][/tex]
3. Eliminate the denominator by multiplying both sides by [tex]\((k+5)(7-k)\)[/tex]:
[tex]\[ 12k - k^2 + 25 = (k+5)(7-k) \][/tex]
4. Expand the right-hand side:
[tex]\[ (k+5)(7-k) = k \cdot 7 + k \cdot (-k) + 5 \cdot 7 + 5 \cdot (-k) = 7k - k^2 + 35 - 5k = 2k - k^2 + 35 \][/tex]
Thus, we have:
[tex]\[ 12k - k^2 + 25 = 2k - k^2 + 35 \][/tex]
5. Isolate the terms involving [tex]\(k\)[/tex] on one side:
[tex]\[ 12k - k^2 + 25 - 2k + k^2 - 35 = 0 \][/tex]
[tex]\[ 10k - 10 = 0 \][/tex]
[tex]\[ 10k = 10 \][/tex]
6. Solve for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{10}{10} \][/tex]
[tex]\[ k = 1 \][/tex]
Therefore, the solution to the equation
[tex]\[ \frac{k}{k+5} + \frac{5}{7-k} = 1 \][/tex]
is [tex]\(\boxed{1}\)[/tex].
[tex]\[ \frac{k}{k+5} + \frac{5}{7-k} = 1 \][/tex]
we will walk through the process in a structured manner.
1. Combine the fractions on the left-hand side:
To do this, we need a common denominator. The common denominator for the fractions [tex]\(\frac{k}{k+5}\)[/tex] and [tex]\(\frac{5}{7-k}\)[/tex] is [tex]\((k+5)(7-k)\)[/tex].
[tex]\[ \frac{k}{k+5} + \frac{5}{7-k} = \frac{k(7-k) + 5(k+5)}{(k+5)(7-k)} \][/tex]
2. Expand and simplify the numerator:
[tex]\[ k(7-k) + 5(k+5) = 7k - k^2 + 5k + 25 = 12k - k^2 + 25 \][/tex]
So, the equation now looks like this:
[tex]\[ \frac{12k - k^2 + 25}{(k+5)(7-k)} = 1 \][/tex]
3. Eliminate the denominator by multiplying both sides by [tex]\((k+5)(7-k)\)[/tex]:
[tex]\[ 12k - k^2 + 25 = (k+5)(7-k) \][/tex]
4. Expand the right-hand side:
[tex]\[ (k+5)(7-k) = k \cdot 7 + k \cdot (-k) + 5 \cdot 7 + 5 \cdot (-k) = 7k - k^2 + 35 - 5k = 2k - k^2 + 35 \][/tex]
Thus, we have:
[tex]\[ 12k - k^2 + 25 = 2k - k^2 + 35 \][/tex]
5. Isolate the terms involving [tex]\(k\)[/tex] on one side:
[tex]\[ 12k - k^2 + 25 - 2k + k^2 - 35 = 0 \][/tex]
[tex]\[ 10k - 10 = 0 \][/tex]
[tex]\[ 10k = 10 \][/tex]
6. Solve for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{10}{10} \][/tex]
[tex]\[ k = 1 \][/tex]
Therefore, the solution to the equation
[tex]\[ \frac{k}{k+5} + \frac{5}{7-k} = 1 \][/tex]
is [tex]\(\boxed{1}\)[/tex].
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