Câu trong ảnh làm như nào các bạn ơi... 2A. Thực hiện phép tính:,$a)~A=\frac1{1.2}+\frac1{2.3}+...+\frac1{99.100};$,$b)~B=(1-\frac12)(1-\frac13)...(1-\frac1{n+1});$,$c)~C=-66.(\frac12-\frac13+\frac1{11})+124.(-37)+63.(-124);$,$d)~D=\frac74(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}).$

Question

Cendela Duong · Accepted Answer

Timi · Answer

Chúng ta sẽ lần lượt giải từng phần trong bài toán.

**Phần a)** Tính $ A = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{99 \cdot 100} $.

Ta có thể sử dụng công thức phân tích như sau:
$$
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
$$
Áp dụng cho $ A $:
$$
A = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{99} - \frac{1}{100} \right)
$$
Như vậy, tất cả các số hạng sẽ bị triệt tiêu, và chúng ta còn lại:
$$
A = 1 - \frac{1}{100} = \frac{99}{100}
$$

**Phần b)** Tính $ B = (1 - \frac{1}{2})(1 - \frac{1}{3}) \ldots (1 - \frac{1}{n+1}) $.

Khi tính từng phần tử, ta có:
$$
1 - \frac{1}{k} = \frac{k-1}{k}
$$
Vậy,
$$
B = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n}{n+1} = \frac{1}{n+1}
$$

**Phần c)** Tính $ C = -66 \cdot \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{11} \right) + 124 \cdot (-37) + 63 \cdot (-124) $.

Đầu tiên, tính phần trong ngoặc:
$$
\frac{1}{2} - \frac{1}{3} + \frac{1}{11} = \frac{3}{6} - \frac{2}{6} + \frac{1}{11} = \frac{1}{6} + \frac{1}{11}
$$
Tính $ \frac{1}{6} + \frac{1}{11} $ theo mẫu số chung 66:
$$
\frac{11}{66} + \frac{6}{66} = \frac{17}{66}
$$
Vậy $ C $ trở thành:
$$
C = -66 \cdot \frac{17}{66} + 124 \cdot (-37) + 63 \cdot (-124)
$$
Rút gọn:
$$
C = -17 + (-4588) + (-7812) = -17 - 4588 - 7812 = -4605 - 7812 = -12417
$$

**Phần d)** Tính $ D = \frac{7}{4} \left( \frac{33}{12} + \frac{3333}{2020} + \frac{333333}{303030} + \frac{33333333}{42424242} \right) $.

Tính từng phần tử trong dấu ngoặc:
1. $ \frac{33}{12} = 2.75 $
2. $ \frac{3333}{2020} = \frac{3333 \div 101}{2020 \div 101} = \frac{33}{20} = 1.65 $
3. $ \frac{333333}{303030} = \frac{333333 \div 111111}{303030 \div 111111} = \frac{3}{3} = 1 $
4. $ \frac{33333333}{42424242} = \frac{33333333 \div 11111111}{42424242 \div 11111111} = \frac{3}{4} = 0.75 $

Cộng tất cả lại:
$$
2.75 + 1.65 + 1 + 0.75 = 6.15
$$

Vậy, tính $ D $:
$$
D = \frac{7}{4} \cdot 6.15 = \frac{7 \cdot 6.15}{4} = \frac{43.05}{4} = 10.7625
$$

Tổng hợp lại:
- $ A = \frac{99}{100} $
- $ B = \frac{1}{n+1} $
- $ C = -12417 $
- $ D = 10.7625 $

Đáp án cuối cùng là:
- $ A = \frac{99}{100} $
- $ B = \frac{1}{n+1} $
- $ C = -12417 $
- $ D = 10.7625 $

Ninh Hoàng · Answer

{"userId":5229300,"userName":"Phạm Tùng Dương tank arena steel battle hack"}

a)

$$A=\frac{1}{1.2}+\frac{1}{2.3}+\cdots+\frac{1}{99.100}$$

$$=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{99}-\frac{1}{100}\right)$$

$$=1-\frac{1}{100}$$

$$=\frac{99}{100}$$

b)

$$B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)\ldots\left(1-\frac{1}{n+1}\right)$$

$$=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\ldots\frac{n}{n+1}$$

$$=\frac{1}{n+1}$$

c)

$$C=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)$$

$$=\left\lbrack\left(-66\right).\frac{1}{2}-\left(-66\right).\frac{1}{3}+\left(-66\right).\frac{1}{11}\right\rbrack+124.\left(-37\right)-63.124$$

$$=\left(-33+22-6\right)+124.\left(-37-63\right)$$

$$=-17+124.\left(-100\right)$$

$$=-17-12400$$

$$=-12417$$

d)

$$D=\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)$$

$$=\frac{7}{4}.\left(\frac{33}{12}+\frac{33.101}{20.101}+\frac{33.10101}{30.10101}+\frac{33.1010101}{42.1010101}\right)$$

$$=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)$$

$$=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)$$

$$=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)$$

$$=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)$$

$$=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)$$

$$=\frac{231}{4}.\frac{4}{21}$$

$$=11$$.

Huycindy · Answer

$a)$
$A = \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dots + \dfrac{1}{99.100}$
$A = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \dots + \left(\dfrac{1}{99} - \dfrac{1}{100}\right)$
$A = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dots + \dfrac{1}{99} - \dfrac{1}{100}$
$A = 1 - \dfrac{1}{100}$
$A = \dfrac{99}{100}$
$b)$
$B = \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\dots\left(1 - \dfrac{1}{n+1}\right)$
$B = \left(\dfrac{2-1}{2}\right) \left(\dfrac{3-1}{3}\right) \dots \left(\dfrac{n+1-1}{n+1}\right)$
$B = \dfrac{1}{2} . \dfrac{2}{3} . \dfrac{3}{4} \dots \dfrac{n}{n+1}$
$B = \dfrac{1 . 2 . 3 \dots n}{2 . 3 . 4 \dots (n+1)}$
$B = \dfrac{1}{n+1}$
$c)$
$C = -66 . \left(\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{11}\right) + 124 . (-37) + 63 . (-124)$
$C = -66 . \dfrac{1}{2} - (-66) . \dfrac{1}{3} + (-66) . \dfrac{1}{11} - 124 . 37 - 124 . 63$
$C = -33 + 22 - 6 - 124 . 37 - 124 . 63$
$C = (-33 + 22 - 6) - 124 . (37 + 63)$
$C = -17 - 124 . 100$
$C = -17 - 12400$
$C = -12417$
$d)$
$D = \dfrac{7}{4} . \left(\dfrac{33}{12} + \dfrac{3333}{2020} + \dfrac{333333}{303030} + \dfrac{33333333}{42424242}\right)$
$D = \dfrac{7}{4} . \left(\dfrac{33:3}{12:3} + \dfrac{3333:101}{2020:101} + \dfrac{333333:30303}{303030:30303} + \dfrac{33333333:3030303}{42424242:3030303}\right)$
$D = \dfrac{7}{4} . \left(\dfrac{11}{4} + \dfrac{33}{20} + \dfrac{11}{10} + \dfrac{11}{14}\right)$
$D = \dfrac{7}{4} . 11 . \left(\dfrac{1}{4} + \dfrac{3}{20} + \dfrac{1}{10} + \dfrac{1}{14}\right)$
$D = \dfrac{77}{4} . \left(\dfrac{5}{20} + \dfrac{3}{20} + \dfrac{2}{20} + \dfrac{1}{14}\right)$
$D = \dfrac{77}{4} . \left(\dfrac{10}{20} + \dfrac{1}{14}\right)$
$D = \dfrac{77}{4} . \left(\dfrac{1}{2} + \dfrac{1}{14}\right)$
$D = \dfrac{77}{4} . \left(\dfrac{7}{14} + \dfrac{1}{14}\right)$
$D = \dfrac{77}{4} . \dfrac{8}{14}$
$D = \dfrac{77}{4} . \dfrac{4}{7}$
$D = \dfrac{77 . 4}{4 . 7}$
$D = 11$

Vu Nguyen · Answer

a)

$$ \begin{align*} A &= \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dots + \dfrac{1}{99.100} \ &= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dots + \dfrac{1}{99} - \dfrac{1}{100} \ &= 1 - \dfrac{1}{100} \ &= \dfrac{99}{100} \end{align*} $$

b)

$$ \begin{align*} B &= \left(1 - \dfrac{1}{2} ight)\left(1 - \dfrac{1}{3} ight) \dots \left(1 - \dfrac{1}{n+1} ight) \ &= \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} \dots \dfrac{n}{n+1} \ &= \dfrac{1}{n+1} \end{align*} $$

c)

$$ \begin{align*} C &= -66 \cdot \left(\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{11} ight) + 124 \cdot (-37) + 63 \cdot (-124) \ &= \left( -66 \cdot \dfrac{1}{2} ight) - \left( -66 \cdot \dfrac{1}{3} ight) + \left( -66 \cdot \dfrac{1}{11} ight) - 124 \cdot 37 - 124 \cdot 63 \ &= -33 + 22 - 6 - 124 \cdot (37 + 63) \ &= -17 - 124 \cdot 100 \ &= -17 - 12400 \ &= -12417 \end{align*} $$

d)

$$ \begin{align*} D &= \dfrac{7}{4} \left( \dfrac{33}{12} + \dfrac{3333}{2020} + \dfrac{333333}{303030} + \dfrac{33333333}{42424242} ight) \ &= \dfrac{7}{4} \left( \dfrac{33}{12} + \dfrac{33 \cdot 101}{20 \cdot 101} + \dfrac{33 \cdot 10101}{30 \cdot 10101} + \dfrac{33 \cdot 1010101}{42 \cdot 1010101} ight) \ &= \dfrac{7}{4} \left( \dfrac{33}{12} + \dfrac{33}{20} + \dfrac{33}{30} + \dfrac{33}{42} ight) \ &= \dfrac{7}{4} \cdot 33 \cdot \left( \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} + \dfrac{1}{5 \cdot 6} + \dfrac{1}{6 \cdot 7} ight) \ &= \dfrac{231}{4} \cdot \left( \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{6} + \dfrac{1}{6} - \dfrac{1}{7} ight) \ &= \dfrac{231}{4} \cdot \left( \dfrac{1}{3} - \dfrac{1}{7} ight) \ &= \dfrac{231}{4} \cdot \dfrac{4}{21} \ &= \dfrac{231}{21} \ &= 11 \end{align*} $$

ft. Hoàng · Answer

a) Ta có: $A = \dfrac{1}{1.2} + \dfrac{1}{2.3} + ... + \dfrac{1}{99.100}$

$= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{99} - \dfrac{1}{100}$

$= 1 - \dfrac{1}{100}$

$= \dfrac{99}{100}$

b) Ta có: $B = \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right).....\left(1 - \dfrac{1}{n+1}\right)$

$= \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} ..... \dfrac{n}{n+1}$

$= \dfrac{1}{n+1}$

c) Ta có: $C = -66.\left(\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{11}\right) + 124.(-37) + 63.(-124)$

$= -66 \cdot \dfrac{1}{2} - (-66) \cdot \dfrac{1}{3} + (-66) \cdot \dfrac{1}{11} - 124 . 37 - 63 . 124$

$= -33 + 22 - 6 - 124.(37 + 63)$

$= -17 - 124 . 100$

$= -17 - 12400$

$= -12417$

d) Ta có: $D = \dfrac{7}{4}\left(\dfrac{33}{12} + \dfrac{3333}{2020} + \dfrac{333333}{303030} + \dfrac{33333333}{42424242}\right)$

$= \dfrac{7}{4}\left(\dfrac{3.11}{12} + \dfrac{33.101}{20.101} + \dfrac{333.1001}{30.1001} + \dfrac{3333.10001}{42.10001}\right)$

$= \dfrac{7}{4}\left(\dfrac{33}{12} + \dfrac{33}{20} + \dfrac{333}{30} + \dfrac{3333}{42}\right)$

$= \dfrac{7}{4}\left(\dfrac{11}{4} + \dfrac{33}{20} + \dfrac{111}{10} + \dfrac{1111}{14}\right)$

$= \dfrac{7}{4} \cdot \dfrac{11}{4} + \dfrac{7}{4} \cdot \dfrac{33}{20} + \dfrac{7}{4} \cdot \dfrac{111}{10} + \dfrac{7}{4} \cdot \dfrac{1111}{14}$

$= \dfrac{77}{16} + \dfrac{231}{80} + \dfrac{777}{40} + \dfrac{1111}{8}$

$= \dfrac{7}{4}\left(\dfrac{385}{140} + \dfrac{231}{140} + \dfrac{1554}{140} + \dfrac{11110}{140}\right)$

$= \dfrac{7}{4} \cdot \dfrac{385 + 231 + 1554 + 11110}{140}$

$= \dfrac{7}{4} \cdot \dfrac{13280}{140}$

$= \dfrac{7}{4} \cdot \dfrac{664}{7}$

$= \dfrac{664}{4}$

$= 166$