How do you simplify \[\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}\]?

Hint: This question is from the topic of pre-calculus. In this question, we are going to use the term iota and the square of iota. We will also use the foil method. In solving this question, we will first rationalize the terms \[\left( 2-5i \right)\] and \[\left( 1-i \right)\]. After rationalizing them, we will multiply the terms which are in the brackets using the foil method. After solving the further equation, we will get our answer.

Complete step by step solution:
Let us solve this question.
The given term which we have to simplify is
\[\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}\]
Let us first rationalize the term \[\left( 2-5i \right)\], we can write
\[\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}\times \dfrac{\left( 2+5i \right)}{\left( 2+5i \right)}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 2+5i \right)}\times \dfrac{\left( 2+5i \right)}{\left( 1-i \right)}\]
Now, let us rationalize the term \[\left( 1-i \right)\], we can write
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 2+5i \right)}\times \dfrac{\left( 2+5i \right)}{\left( 1-i \right)}\times \dfrac{\left( 1+i \right)}{\left( 1+i \right)}\]
As we know that \[\left( a-b \right)\left( a+b \right)\] can also be written as \[\left( {{a}^{2}}-{{b}^{2}} \right)\], so we can write the above equation as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( {{2}^{2}}-{{\left( 5i \right)}^{2}} \right)}\times \dfrac{\left( 2+5i \right)\times \left( 1+i \right)}{\left( {{1}^{2}}-{{\left( i \right)}^{2}} \right)}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 4-25{{i}^{2}} \right)}\times \dfrac{\left( 2+5i \right)\left( 1+i \right)}{\left( 1-{{i}^{2}} \right)}\]
As we know that \[{{i}^{2}}=-1\], so we can write the above equation as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 4-25\left( -1 \right) \right)}\times \dfrac{\left( 2+5i \right)\left( 1+i \right)}{\left( 1-\left( -1 \right) \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 4+25 \right)}\times \dfrac{\left( 2+5i \right)\left( 1+i \right)}{\left( 1+1 \right)}\]
The equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{\left( 5+i \right)\left( 4-3i \right)}{29}\times \dfrac{\left( 2+5i \right)\left( 1+i \right)}{2}\]
Now, using the foil method, let us multiply the terms.
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{5\times 4+5\times \left( -3i \right)+i\times 4+i\times \left( -3i \right)}{29}\times \dfrac{2\times 1+2\times i+5i\times 1+5i\times i}{2}\]
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{20+\left( -15i \right)+4i+\left( -3{{i}^{2}} \right)}{29}\times \dfrac{2+2i+5i+5{{i}^{2}}}{2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{20-11i+\left( -3\left( -1 \right) \right)}{29}\times \dfrac{2+7i+5\left( -1 \right)}{2}=\dfrac{20-11i+3}{29}\times \dfrac{2+7i-5}{2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{23-11i}{29}\times \dfrac{-3+7i}{2}=\dfrac{\left( 23-11i \right)\left( -3+7i \right)}{29\times 2}\]
Again using foil method, we can write the above equation as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{23\times \left( -3 \right)+23\times 7i+\left( -11i \right)\times \left( -3 \right)+\left( -11i \right)\times 7i}{29\times 2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{-69+161i+33i-77{{i}^{2}}}{29\times 2}=\dfrac{-69+194i+77}{29\times 2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{8+194i}{29\times 2}=\dfrac{4+97i}{29}\]
The above equation can also be written as
\[\Rightarrow \dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}=\dfrac{4}{29}+\dfrac{97i}{29}\]

Hence, we have simplified the term \[\dfrac{\left( 5+i \right)\left( 4-3i \right)}{\left( 2-5i \right)\left( 1-i \right)}\] and the simplified value is \[\dfrac{4}{29}+\dfrac{97}{29}i\]

Note: We should have a better knowledge in the topic of pre-calculus and complex numbers. We should know about the term iota. The symbol for the term iota is \[i\] and it is equal to the square root of negative one. Or, in mathematical term we can that
\[i=\sqrt{-1}\]
\[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1\]
The formula for foil method is
\[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\]
We should also know about rationalization.
Whenever we have to rationalize a complex number like \[a+ib\], then we will multiply the term \[a-ib\] to both the numerator and denominator.