E=0,5 x 199/200=199/400

\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)

A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0

A = 0*

*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)

a) \(\left[\left(-2,7\right)^4\right]^5-\left[\left(-2,7\right)^2\right]^{20}\)

\(=\left(-2,7\right)^{20}-\left(-2,7\right)^{20}\)

\(=0\)

b) \(\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\dfrac{17}{2}\right)^7:\left(\dfrac{17}{2}\right)^6\)

\(=\left(-0,5\right)^2-\dfrac{17}{2}\)

\(=0,25-\dfrac{17}{2}\)

\(=-8,25\)

c) \(\left(8^{14}:4^{12}\right):\left(16^6:8^2\right)\)

\(=8^{14}:4^{12}:16^6\cdot8^2\)

\(=2^{48}:2^{24}:2^{24}\)

\(=0\)

Cái cuối bằng 1 nhé!

Sửa lại cái cuối cùng đáp án bằng 0 thôi, còn cách giải chi tiết mình làm đúng rồi nhé. Bởi:

\(2^{48}:2^{24}:2^{24}=2^0=1\)

a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)

=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)

=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)

=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)

=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)

=\(\dfrac{36}{208}-\dfrac{5}{3}\)

=\(\dfrac{108}{624}-\dfrac{1040}{624}\)

=\(\dfrac{-932}{624}\)

=\(\dfrac{233}{156}\)

                                 còn câu b mk chưa học nên mk chịu

Giải:

5/9:13/7+5/9:13/9 -1 2/3

=5/9.7/13+5/9.9/13-5/3

=5/9.(7/13+9/13)-5/3

=5/9.16/13-5/3

=80/117-5/3

=-115/117

4 2/5 : 0,5% -1 3/7 .14% +(-0,5)

=22/5:1/200-10/7.7/50 +(-1/2)

=880-1/5-1/2

=8793/10

\(1\dfrac{1}{6}.2^2-0,5:\dfrac{3}{10}+\dfrac{1}{3}\)

\(=\dfrac{7}{6}.4-\dfrac{1}{2}.\dfrac{10}{3}+\dfrac{1}{3}\)

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

\(=\dfrac{14}{3}-\dfrac{5}{3}+\dfrac{1}{3}\)

\(=\dfrac{14-5+1}{3}\)

\(=\dfrac{10}{3}\)

\(\Leftrightarrow\left(x-0.5\right)\cdot\dfrac{-4}{x-0.5}=-1\cdot\left(-4\right)\)

=>-4=4(loại)

\(1\dfrac{13}{15}\cdot\left(0,5\right)^2+3\cdot\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4}+3\cdot\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{7}{15}+3\cdot\dfrac{37}{20}\cdot\dfrac{24}{47}\)

\(=\dfrac{7}{15}+\dfrac{666}{235}=\dfrac{2327}{705}\)

\(1\dfrac{13}{15}.\left(0,5\right)^2+3.\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\) 

\(=\dfrac{28}{15}.\dfrac{1}{4}+3.\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\) 

\(=\dfrac{7}{15}+3.\dfrac{37}{20}:\dfrac{47}{24}\) 

\(=\dfrac{7}{15}+\dfrac{666}{235}\) 

\(=\dfrac{2327}{705}\)

\(M=1\dfrac{1}{10}\cdot\dfrac{15}{19}-\dfrac{22}{38}\cdot0,5+110\%\cdot\dfrac{9}{19}\)

\(=\dfrac{11}{10}\cdot\dfrac{15}{19}-\dfrac{11}{19}\cdot\dfrac{1}{2}+\dfrac{11}{10}\cdot\dfrac{9}{19}\)

\(=\dfrac{15}{10}\cdot\dfrac{11}{19}-\dfrac{11}{19}\cdot\dfrac{1}{2}+\dfrac{9}{10}\cdot\dfrac{11}{19}\)

\(=\dfrac{11}{19}\left(\dfrac{15}{10}-\dfrac{1}{2}+\dfrac{9}{10}\right)\\ =\dfrac{11}{19}\cdot\dfrac{19}{10}\\ =\dfrac{11}{10}\)