a) 15/11 - (5/7 - 18/11) + 27/7

= 15/11 - 5/7 + 18/11 + 27/7

= (15/11 + 18/11) + (-5/7 + 27/7)

= 3 + 22/7

= 43/7

b) 39/5 + (9/4 - 9/5) - (5/4 + 1,2)

= 39/5 + 9/4 - 9/5 - 5/4 - 6/5

= (39/5 - 9/5 - 6/5) + (9/4 - 5/4)

= 24/5 + 1

= 29/5

c) -1,2 - 0,8 + 0,25 + 5,75 - 2022

= (-1,2 - 0,8) + (0,25 + 5,76) - 2022

= -2 + 6 - 2022

= 4 - 2022

= -2018

d) 0,1 + 16/9 + 5,1 + (-20/9)

= (0,1 + 5,1) + (16/9 - 20/9)

= 5,2 - 4/9

= 419/90

a) \(\dfrac{15}{11}-\left(\dfrac{5}{7}-\dfrac{18}{11}\right)+\dfrac{27}{7}=\dfrac{22}{7}+3=\dfrac{43}{77}\)

b) \(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{5}\right)=\dfrac{24}{5}+1=\dfrac{29}{5}\)

c) \(-1,2-0,8+0,25+5,75-2022=-2+6-2022=-2018\)

d) \(0,1+\dfrac{16}{9}+5,1+\dfrac{-20}{9}=\dfrac{26}{5}-\dfrac{4}{9}=\dfrac{214}{45}\)

Bạn xem bài tương tự tại đây. Đề là:
Tính $(1+\frac{1}{1.3})(1+\frac{1}{2.4})....(1+\frac{1}{2021.2023})$

\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)

dỗiii c-các cậu vì ko lm hộ t-tớ -((

\(-2x^2=-32\)

\(\Rightarrow x^2=\dfrac{-32}{-2}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Vậy: ... 

\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)

\(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)

\(=\left(3-5-6\right)-\left(\dfrac{1}{4}-\dfrac{7}{4}\right)-\dfrac{3}{2}+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\dfrac{6}{5}\)

\(=-8-\dfrac{6}{4}-\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{6}{5}\)

\(=-8-\dfrac{3}{2}-\dfrac{3}{2}+1+\dfrac{6}{5}\)

\(=-7-3+\dfrac{6}{5}\)

\(=\dfrac{6}{5}-10\)

\(=-\dfrac{44}{5}\)

Để A = 0 thì |x² - 1| = 0 và |x + 1| = 0

*) |x² - 1| = 0

x² - 1 = 0

x² = 1

x = 1 hoặc x = -1 (1)

*) |x + 1| = 0

x + 1 = 0

x = -1 (2)

Từ (1) và (2) ⇒ x = -1

Vậy x = -1 thì A = 0

Bạn xem lại nhé mình trả lời rồi nha bạn không nên đăng lại nhé ! 

  Qua C vẽ tia Cz // DE

⇒ ∠DCz = ∠CDE = 108⁰

⇒ ∠zCA = ∠DCz - ∠DCA

= 108⁰ - 48⁰

= 60⁰

⇒ ∠zCA = ∠CAB = 60⁰

Mà ∠zCA và ∠CAB là hai góc so le trong

⇒ Cz // AB

Lại có:

Cz // DE

⇒ AB // DE