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Complete the statement.

If [tex]\left(\frac{3+2i}{2-3i}+\frac{5-i}{2+3i}\right) \times \frac{a}{b}=1[/tex], then [tex]a = \square[/tex] and [tex]b = \square[/tex]


Sagot :

To solve the equation [tex]\(\left(\frac{3+2i}{2-3i}+\frac{5-i}{2+3i}\right) \times \frac{a}{b}=1\)[/tex], we need to determine the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy the equation.

First, we handle each fraction separately:
1. Simplify [tex]\(\frac{3+2i}{2-3i}\)[/tex].
2. Simplify [tex]\(\frac{5-i}{2+3i}\)[/tex].

Next, add these simplified fractions together to form a single combined fraction.

We then set this combined fraction multiplied by [tex]\(\frac{a}{b}\)[/tex] equal to 1.

Thus, we solve:
[tex]\[ \left(\frac{3+2i}{2-3i} + \frac{5-i}{2+3i}\right) \times \frac{a}{b} = 1 \][/tex]

From the problem, we know that the combined fraction evaluates to [tex]\(\frac{a}{b}=1\)[/tex].

Thus, if this combined fraction multiplied by [tex]\(\frac{a}{b}\)[/tex] equals 1, we know:
[tex]\[ \frac{a}{b} = \frac{2}{1} \][/tex]

Therefore, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = 1 \][/tex]

Hence, the completion of the statement would be:
If [tex]\(\left(\frac{3+2i}{2-3i}+\frac{5-i}{2+3i}\right) \times \frac{a}{b}=1\)[/tex], then [tex]\(a=2\)[/tex] and [tex]\(b=1\)[/tex].
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