Lời giải:
$A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{1998}{1999}=\frac{1.2.3....1998}{2.3.4...1999}=\frac{1}{1999}$

tự lm haha

giải thích nhé

`#3107.101107`

Đặt $A = 1 + 2 + 2^2 + 2^3 + ... + 2^{50}$

$2A = 2 + 2^2 + 2^3 + ... + 2^{51}$

$2A - A = (2 + 2^2 + 2^3 + ... + 2^{51}) - (1 + 2 + 2^2 + ... + 2^{50})$

$A = 2 + 2^2 + 2^3 + ... + 2^{51] - 1 - 2 - 2^2 - ... - 2^{50}$

$A = 2^{51} - 1$

Vậy, `A =` $2^{51} - 1.$

\(1\dfrac{2}{3}x+x=\left(-1,6\right)\\ \Rightarrow\dfrac{5}{3}x+x=\dfrac{18}{5}\\ \Rightarrow\dfrac{8}{3}x=\dfrac{18}{5}\\ \Rightarrow x=\dfrac{20}{27}\)

Sửa:

\(1\dfrac{2}{3}x+x=\left(-1,6\right)\\ \Rightarrow\dfrac{5}{3}x+x=-\dfrac{8}{5}\\ \Rightarrow\dfrac{8}{3}x=-\dfrac{8}{5}\\ \Rightarrow x=-\dfrac{20}{27}\)

a) Thể tích thùng xe tải:

5,6 . 2 . 2 = 22,4 (m³)

b) Thể tích gói hàng:

0,5³ = 0,125 (m³)

Số gói hàng tối đa thùng xe có thể chở:

22,4 : 0,125 = 179,2 (gói) ≈ 179 (gói)

chỉ với

\(-\dfrac{3}{7}.\dfrac{5}{7}+\left(-\dfrac{5}{7}\right).\dfrac{8}{11}+\dfrac{19}{7}\\ =\dfrac{5}{7}.\left(-\dfrac{3}{7}-\dfrac{8}{11}\right)+\dfrac{19}{7}\\ =\dfrac{5}{7}.-\dfrac{89}{77}+\dfrac{19}{7}\\ =\dfrac{445}{539}+\dfrac{19}{7}\\ =\dfrac{1908}{539}\)

`#3107.101107`

`x^2 - 0,04 = 0`

`\Rightarrow x^2 = 0,04`

`\Rightarrow x^2 = (+-0,2)^2`

`\Rightarrow x = +- 0,2`

Vậy, `x \in {-0,2; 0,2}.`

`#3107.101107`

c)

$-x + \dfrac{3}2 = x + \dfrac{3}5$

$\Rightarrow -x - x = \dfrac{3}5 - \dfrac{3}2$

$\Rightarrow -2x = -\dfrac{9}{10}$

$\Rightarrow 2x = \dfrac{9}{10}$

$\Rightarrow x = \dfrac{9}{10} \div 2$

$\Rightarrow x = \dfrac{9}{20}$

Vậy, $x = \dfrac{9}{20}.$

Ta có:

∠COD + ∠COB + ∠BOA = 180⁰

⇒ ∠COB = 180⁰ - (∠COD + ∠BOA)

= 180⁰ - (30⁰ + 20⁰)

= 180⁰ - 50⁰

= 130⁰