We know that for problem solving for algebra expressions must follow 'BODMAS':

B - Brackets

O - Order

D - Division

M - Multiplication

A - Addition

S - Subtraction

Step 1: Expand the terms in the brackets first

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Expand}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

Step 2: Multiply the expanded terms

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Expand}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

\(\begin{aligned} &=9k^2-1+\dfrac{9(k^2+12k+36)}{2}\\ &=9k^2-1+\dfrac{9}{2}k^2+54k+162\\ \end{aligned}\)

Step 3: Solve the same terms

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Expand}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

\(\begin{aligned} &=9k^2-1+\dfrac{9(k^2+12k+36)}{2}\\ &=9k^2-1+\dfrac{9}{2}k^2+54k+162\\&=9k^2+\dfrac{9}{2}k^2+54k-1+162\\ &=\dfrac{27}{2}k^2+54k+161. \end{aligned}\)

Step 4: Make a conclusion

Therefore,

\((3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\ =\dfrac{27}{2}k^2+54k+161.\)

Kita mengetahui bahawa penyelesaian masalah untuk ungkapan algebra mestilah mengikut 'BODMAS':

B - Brackets

O - Order

D - Division

M - Multiplication

A - Addition

S - Subtraction

Langkah 1: Mengembangkan sebutan di dalam kurungan

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Kembangkan}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

Langkah 2: Mendarabkan sebutan yang dikembangkan

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Kembangkan}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

\(\begin{aligned} &=9k^2-1+\dfrac{9(k^2+12k+36)}{2}\\ &=9k^2-1+\dfrac{9}{2}k^2+54k+162\\ \end{aligned}\)

Langkah 3: Menyelesaikan sebutan yang sama

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Kembangkan}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

\(\begin{aligned} &=9k^2-1+\dfrac{9(k^2+12k+36)}{2}\\ &=9k^2-1+\dfrac{9}{2}k^2+54k+162\\&=9k^2+\dfrac{9}{2}k^2+54k-1+162\\ &=\dfrac{27}{2}k^2+54k+161. \end{aligned}\)

Langkah 4: Membuat kesimpulan

Maka,

\((3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\ =\dfrac{27}{2}k^2+54k+161.\)

We know that for problem solving for algebra expressions must follow 'BODMAS':

B - Brackets

O - Order

D - Division

M - Multiplication

A - Addition

S - Subtraction

Step 1: Expand the terms in the brackets first

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Expand}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

Step 2: Multiply the expanded terms

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Expand}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

\(\begin{aligned} &=9k^2-1+\dfrac{9(k^2+12k+36)}{2}\\ &=9k^2-1+\dfrac{9}{2}k^2+54k+162\\ \end{aligned}\)

Step 3: Solve the same terms

\(\begin{aligned} &(3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k+6)(k+6)}{2}\longrightarrow\text{Expand}\\\\ &=[(3k)(3k)+(3k)(1)\\&\quad+(-1)(3k)+(-1)(1)]\\ &\quad+\dfrac{9[(k)(k)+(k)(6)+(6)(k)+(6)(6)]}{2}\\\\ &=9k^2+3k-3k-1\\ &\quad+\dfrac{9(k^2+6k+6k+36)}{2} \end{aligned}\)

\(\begin{aligned} &=9k^2-1+\dfrac{9(k^2+12k+36)}{2}\\ &=9k^2-1+\dfrac{9}{2}k^2+54k+162\\&=9k^2+\dfrac{9}{2}k^2+54k-1+162\\ &=\dfrac{27}{2}k^2+54k+161. \end{aligned}\)

Step 4: Make a conclusion

Therefore,

\((3k-1)(3k+1)+\dfrac{9(k+6)^2}{2}\\ =\dfrac{27}{2}k^2+54k+161.\)