Question

Simplify :
1.[{(25/5-5/5)}]/[{(5/5+2/2)}].​

Answer 1

Answer:

(5-1) /(1+1)

4/2

2

Step-by-step explanation:

therefore answer is 2

nice sum

hope it helps u

Answer 2

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{darkblue}{Simplify:}}}}}}}\end{gathered}$

$\begin{gathered} \dashrightarrow \sf{ \frac{ \dfrac{25}{5} - \dfrac{5}{5} }{ \dfrac{5}{5} + \dfrac{2}{2} }} \end{gathered}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{darkblue}{Solution:}}}}}}}\end{gathered}$

$\begin{gathered} \dashrightarrow \sf{ \frac{ \dfrac{25}{5} - \dfrac{5}{5} }{ \dfrac{5}{5} + \dfrac{2}{2} }} \end{gathered}$

$\begin{gathered} \dashrightarrow \sf{ \frac{{\cancel{\dfrac{25}{5}}} - {\cancel{\dfrac{5}{5}}}}{{\cancel{\dfrac{5}{5}}} +{\cancel{\dfrac{2}{2}}}}} \end{gathered}$

$\begin{gathered}\dashrightarrow \sf{\dfrac{5 - 1}{1 + 1}} \end{gathered}$

$\begin{gathered}\dashrightarrow \sf{\dfrac{4}{2}} \end{gathered}$

$\begin{gathered}\dashrightarrow \sf{\cancel{\dfrac{4}{2}}}\end{gathered}$

$\begin{gathered}\dashrightarrow \sf{2}\end{gathered}$

$\begin{gathered}\dashrightarrow {\underline{\boxed{\sf{\pink{ \: \: \: 2 \: \: \: }}}}}\end{gathered}$

• Henceforth,The Answer is 2.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{darkblue}{Additional Information:}}}}}}}\end{gathered}$

$\begin{gathered} \tiny{\boxed{\begin{array}{cc} \sf{(a + b)^{2}  = a^{2} + 2ab + b^{2}} \\ \\ \sf{(a - b)^{2} = a^{2} - 2ab + b^{2}} \\  \\ \sf{a^2 - b^2 = (a - b)(a + b)} \\ \\   \sf{(x + a)(x + b) = x^2 + x(a + b) + ab} \\ \\  \sf{(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)}  \\ \\  \sf{(a + b)^3 = a^3 + b^3 + 3ab(a + b)} \\  \\ \sf{(a - b)^3 = a^3 - b^3 - 3ab(a - b)}  \\  \\ \sf{a^3 + b^3 = (a + b)(a^2 - ab + b^2)} \\  \\  \sf{a^3 - b^3 = (a - b)(a^2 + ab + b^2)}  \\ \\  \sf{a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)} \\ \\ \sf{a^n \times a^b} = a^{n+b} \\   \\ \sf{a^n \times b^n= ab^n}  \\ \\  \sf{a^{-n} = 1/a^n}  \\ \\ \sf{a^n \div a^b = a^{n-b}}  \\ \\  \sf{(a^n/b^n) = (a/b)^n} \\   \\\sf{a^0 = 1} \\\end{array}}}\end{gathered}$