giúp tớ bài 9 với
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Đây là toán nâng cao của nâng cao chuyên đề dãy số cách đều, cấu trúc
Giải:
Cứ 3 lon bia đổi được 1 lon bia nên số lon bia mất đi sau mỗi lần đổi là:
3 - 1 = 2 (lon bia)
Sau lần đổi thứ nhất số lon bia còn lại là: 30 - 2 = 28 (lon)
Sau lần đổi cuối cùng số lon bia còn lại là 2 lon (vì 2 < 3 nên không thể đổi được nữa)
Số lần đổi vỏ lon bia là: (28 - 2) : 2 + 1 = 14 (lần)
Vậy tổng số lon bia mà ngườ đó có thể uống được khi mua 30 lon bia và được tặng là:
3 x 14 + 2 = 44 (lon bia)
Đáp số: 44 lon bia

a: ΔABC có BH\(\perp\)AC
nên \(S_{ABC}=\dfrac{1}{2}\cdot BH\cdot AC\)
ΔACB có CK\(\perp\)AB
nên \(S_{ABC}=\dfrac{1}{2}\cdot CK\cdot AB\)
b: \(S_{ABC}=\dfrac{1}{2}\cdot CK\cdot AB\)
\(S_{ABC}=\dfrac{1}{2}\cdot BH\cdot AC\)
Do đó: \(CK\cdot AB=BH\cdot AC\)
mà AB=AC
nên BH/CK=1
c: Vì BH/CK=1
nên BH=CK

A.
1. How far is it from Vinh to Ha Noi City?
2. I used to walk to school when I was a student.
3. In spite of their shortness, they still love playing sports.
4. Although she is poor, everybody has a great regard for her.
5. Although Tien Dung goalkeeper is so young, he performs excellently.
6. In spite of their old age, they worked very hard.
7. Because of Peter's sickness, he didn't go to school yesterday.
B.
1. How far is it from your house to school?
2. Where did you go with your friends yesterday?
3. What subject did he teach in the high school?
4. Who did you see last night?
A
1 How far is it from Vinh to Ha Noi city?
2 I used to walk to school when I was a student
3 In spite of being short, they still love playing sports
4 Although she was poor, everybody has a great regard for her
6 Althugh Tien Dung goalkeeper was so young, he performs excellently
7 In spite of being old, they worked very hard
8 Because of being sick, Peter didn't go to school yesterday


Question V:
1. goes
2. go
3. seeing
4. to play
5. has learning
6. watched
7. will visit
8. working
Question VI:
1. celebration
2. festive
3. exciting
4. amazed
V
1 goes
2 go
3 seeing
4 to play
5 has learned
6 watched
7 is going to visit
8 working
Bài 9:
\(A=A_1A_2A_3...A_{100}=(\frac{1}{2}x)(\frac{2}{3}x^2)(\frac{3}{4}x^3)...(\frac{100}{101}x^{100})\)
\(=(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{100}{101})(x.x^2.x^3...x^{100})\\ =\frac{1.2.3...100}{2.3.4...100.101}.x^{1+2+3+...+100}\\ =\frac{1}{101}.x^{100.101:2}=\frac{x^{5050}}{101}\)