\(\widehat{B_1}\) = \(\widehat{B_2}\) = 100⁰ (hai góc đối đỉnh)

\(\widehat{C_2}\) = \(\widehat{B_1}\) = 100⁰ (hai góc đồng vị)

\(\widehat{C_3}\) + \(\widehat{C_2}\) = 180⁰ ⇒ \(\widehat{C_3}\) = 180⁰ - 100⁰ = 80⁰

\(\widehat{D_1}\) = \(\widehat{A_1}\) = 60⁰ (so le trong) 

\(\widehat{DAH}\) = 90⁰ - 60⁰ = 30⁰

Mik ns mấy cái số tại mik sợ nó bị mờ nhoa

a) \(-\dfrac{3}{5}-x=-0,75\)

\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)

b) \(1\dfrac{4}{5}=-0,15-x\)

\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)

\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)

\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)

c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)

a) \(-\dfrac{3}{5}-x=-0,75\)

\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)

\(x=\dfrac{27}{20}\)

________

b) \(1\dfrac{4}{5}=-0,15-x\)

\(=>-0,15-x=\dfrac{9}{5}\)

\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)

\(x=\dfrac{-39}{20}\)

c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)

\(x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)

\(x=\dfrac{6}{15}=\dfrac{2}{5}\)

\(a,0,\left(123\right)+0,\left(876\right)=\dfrac{123}{999}+\dfrac{876}{999}=\dfrac{999}{999}=1\left(đpcm\right)\\ b,0,\left(123\right).3+0,\left(630\right)=\dfrac{123}{999}.3+\dfrac{630}{999}=\dfrac{369}{999}+\dfrac{630}{999}=\dfrac{999}{999}=1\left(đpcm\right)\)

Đổ vào bể số nước là:

30 x 120 = 3600 (L nước)

               = 3,6m³

Chiều rộng của bể nước là:

3,6 : 0,6 : 3 = 2 (m)

Đáp số: 2m

\(3,\left(5\right)=\dfrac{32}{9}\)

\(3,\left(5\right)=3+0,\left(5\right)=3+5.0,\left(1\right)=3+5.\dfrac{1}{9}=3+\dfrac{5}{9}=\dfrac{32}{9}\)

đáp án ....... ...¿.¿¿¿

\(\left(-39,82\right):\left(-7,24\right)\)

\(=39,82:7,24\)

\(=5,5\)

5,5 nha

a) A = 1/(1.2) + 1/(2.3) + ... + 1/[n(n + 1)]

= 1 - 1/2 + 1/2 - 1/3 + 1/n - 1/(n + 1)

= 1 - 1/(n + 1)

b) Do n ∈ ℕ

⇒ n + 1 > 0

⇒ 1/(n + 1) > 0

⇒ 1 - 1/(n + 1) < 1

Vậy A < 1

b3:

a, <

b, <

c, <

d, >

c2, >

f, >

Bài 4:

0,4636363... < 0,463736< 0,4656365< 0,466< 7/15

b3:

a, <

b, <

c, <

d, >

c2, >

f, >

Bài 4:

0,4636363... < 0,463736< 0,4656365< 0,466< 7/15