Rewrite each expression in simplest form:
\[\begin{align}
  & \left( a \right)6\times x\times y \\
 & \left( b \right)2\times y\times y \\
 & \left( c \right)5\times b\times a \\
 & \left( d \right)4x\div 2y \\
 & \left( e \right)4\times x+5\times y \\
 & \left( f \right)3\times \left( x+1 \right)+2\times x \\
 & \left( g \right)7\times a\times b \\
 & \left( h \right)a\times 4\times b \\
 & \left( i \right)y\times z\times z \\
 & \left( j \right)\left( x+3 \right)\div 4 \\
 & \left( k \right)a\times 7-2\times b \\
 & \left( l \right)2\times \left( x+4 \right)\div 3 \\
\end{align}\]

Hint: In this question, we are given various expressions and we need to simplify all of them. For this, we just need to remove multiplication and division signs. Also, we need the constant to be written first and then the variables. All like terms should be added. In case of the same variable being multiplied two or more times, we just need to raise its power and write the variable one time only.

Complete step by step answer:
Let us solve the given expression one by one.
 $ \left( a \right)6\times x\times y $ .
All are unlike terms and in multiplication form, they can be written together. So removing the multiplication sign we get 6xy which is our required answer.
 $ \left( b \right)2\times y\times y $ .
y is repeated two times so it can be written as $ {{y}^{2}} $ . Removing the multiplication sign between 2 and $ {{y}^{2}} $ we get $ 2{{y}^{2}} $ which is our required answer.
 $ \left( c \right)5\times b\times a $ .
All are unlike, so simply removing the multiplication sign we get 5ba which is our required answer.
 $ \left( d \right)4x\div 2y $ .
2y can be written in the denominator so we get $ \dfrac{4x}{2y} $ . Now 4 is divided by 2 so after dividing we get $ \dfrac{2x}{y} $ which is our required answer.
 $ \left( e \right)4\times x+5\times y $ .
Removing the multiplication sign between 4 and x, 5 and y we get 4x+5y. It cannot be further simplified so this is our required answer.
 $ \left( f \right)3\times \left( x+1 \right)+2\times x $ .
Removing multiplication signs we get $ 3\left( x+1 \right)+2x $ opening brackets we get $ 3x+3+2x $ . 3x and 2x are like terms, so adding them we get 5x+3 which is our required answer.
 $ \left( g \right)7\times a\times b $ .
Removing multiplication signs we get 7ab which is our required answer.
 $ \left( h \right)a\times 4\times b $ .
Removing multiplication we get a4b but constant should always be written first. So rearranging we get, 4ab which is our required answer.
 $ \left( i \right)y\times z\times z $ .
Here z is multiplied with z so raising the powers of z to 2 and removing multiplication signs we get, $ y{{z}^{2}} $ which is our required answer.
 $ \left( j \right)\left( x+3 \right)\div 4 $ .
Removing the division sign by writing 4 in the denominator we get, $ \dfrac{\left( x+3 \right)}{4} $ .
It cannot be simplified further so this is our required answer.
 $ \left( k \right)a\times 7-2\times b $ .
Removing multiplication sign between a and 7, 2 and b we get a7-2b. It cannot be simplified further. So this is our final answer.
 $ \left( l \right)2\times \left( x+4 \right)\div 3 $ .
Removing the division sign by writing 3 in the denominator we get $ \dfrac{2\times \left( x+4 \right)}{3} $ .
Removing the multiplication sign in the numerator we get $ \dfrac{2\left( x+4 \right)}{3} $ .
Opening bracket we get $ \dfrac{2x+8}{3} $ which is our required answer.

Note:
Students should solve each expression step by step without forgetting any terms. Make sure to write constantly in terms. Two variables written together are just assumed as multiplied together which is why we have removed the signs.