0,94

Xét tam giác SOI có OI=2a,OH=căn2 a

=> SO= 2a

=> cos(1/căn2)=45°

Đặt x=5(triệu đồng), y=0,33%

5 năm=60 tháng

Gọi \(A_{n}\) là số tiền ông Đại có được sau n tháng

\(A_{1} = x \left(\right. 1 + y \left.\right) \left(\right. đ \overset{ˋ}{\hat{o}} n g \left.\right)\)

\(A_{2} = \left(\right. A_{1} + x \left.\right) \cdot \left(\right. 1 + y \left.\right) = \left[\right. x \left(\right. 1 + y \left.\right) + x \left]\right. \left(\right. 1 + y \left.\right) = x \left(\left(\right. 1 + y \left.\right)\right)^{2} + x \left(\right. 1 + y \left.\right) \left(\right. đ \overset{ˋ}{\hat{o}} n g \left.\right)\)

\(A_{3} = \left(\right. A_{2} + x \left.\right) \left(\right. 1 + y \left.\right) = x \left(\left(\right. 1 + y \left.\right)\right)^{3} + x \left(\left(\right. 1 + y \left.\right)\right)^{2} + x \left(\right. 1 + y \left.\right)\)(đồng)

...

\(A_{n} = \left(\right. A_{n - 1} + x \left.\right) \left(\right. 1 + y \left.\right) = x \left(\left(\right. 1 + y \left.\right)\right)^{n} + x \left(\left(\right. 1 + y \left.\right)\right)^{n - 1} + . . . + x \left(\right. 1 + y \left.\right) \left(\right. đ \overset{ˋ}{\hat{o}} n g \left.\right)\)

\(= x \left(\right. 1 + y \left.\right) \cdot \frac{1 - \left(\left(\right. 1 + y \left.\right)\right)^{n}}{1 - \left(\right. 1 + y \left.\right)}\)

=>\(A_{60} = 5 \left(\right. 1 + 0 , 33 \% \left.\right) \cdot \frac{1 - \left(\left(\right. 1 + 0 , 33 \% \left.\right)\right)^{60}}{1 - \left(\right. 1 + 0 , 33 \% \left.\right)} \simeq 332 , 25 \left(\right. t r i ệ u đ \overset{ˋ}{\hat{o}} n g \left.\right)\)