so tien ban dau: 200 trieu

lai suat thang: r = 0.0045

thoi gian: 4 nam = 48 thang

rut moi thang so tien x

day la bai toan rut tien deu (cong thuc annuity):

200 = x * (1 - (1 + r)^(-48)) / r

=> x = 200 * r / (1 - (1 + r)^(-48))

thay r = 0.0045:

x ≈ 200 * 0.0045 / (1 - (1.0045)^(-48))

tinh gan dung:

(1.0045)^48 ≈ 1.24

=> (1.0045)^(-48) ≈ 1 / 1.24 ≈ 0.806

=> mau so ≈ 1 - 0.806 = 0.194

=> x ≈ 0.9 / 0.194 ≈ 4.64

dap an: khoang 4.6 trieu moi thang

day la hinh vuong, bd = 2a => canh hinh vuong la a can 2

goi chieu cao h

goc giua mat phang (a’bd) va (abcd) = 30 do

xet mat phang chua duong cao, ta co:

tan 30 do = h / (bd/2)

ma bd = 2a => bd/2 = a

nen:

tan 30 = h / a => h = a / can 3

khoang cach tu a den mat phang (a’bd) la:

d = h * sin 30 = (a / can 3) * (1/2) = a / (2 can 3)

dap an: a / (2 can 3)

goi canh lap phuong la a. theo de bai ab = 4 can 5 nen a = 4 can 5

dat he truc toa do:

a(0,a,0), b(a,a,0)

tam mat doi dien la o(a/2, a/2, a)

vec to ab = (a,0,0)

vec to ao = (a/2, -a/2, a)

khoang cach tu diem o den duong thang ab:

d = do dai cua (ao x ab) chia cho do dai ab

tinh tich co huong:

ao x ab = (0, a^2, a^2/2)

do dai:

|ao x ab| = a^2 * can(5) / 2

|ab| = a

suy ra:

d = (a^2 * can5 / 2) / a = a * can5 / 2

thay a = 4 can 5:

d = (4 can 5 * can 5) / 2 = 20 / 2 = 10

dap an: 10