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a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
1/
Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.
Số số hạng: $(101-1):4+1=26$
$A=(101+1)\times 26:2=1326$
2/
$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$
$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$
$=(1+2+2^2)(1+2^3+2^6+2^9)$
$=7(1+2^3+2^6+2^9)\vdots 7$
b: \(27D=3^{14}+3^{17}+...+3^{2024}\)
\(\Leftrightarrow26D=3^{2024}-3^{11}\)
hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)
c: \(25E=-5^4-5^6-...-5^{1002}\)
\(\Leftrightarrow24E=-5^{1002}+5^2\)
hay \(E=\dfrac{-5^{1002}+5^2}{24}\)
A=[(1+9)+(3+7)+(5+15)+(11+13)].2
A=[10+10+20+24].2
A=64.2
A=128
B=(17+23)+(19+21)+(23+27)+(25+29)
B=40+40+50+54
B=tự tính
A=(1+9)+(3+7)+(5+15)+(7+13)+(11.2)
A=10+10+10+10+22
A=40+22
A=62
mình làm câu A thôi nhé
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
\(\dfrac{-2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{19.21}-\dfrac{2}{21.23}-\dfrac{2}{23.25}-\dfrac{2}{25.27}-\dfrac{1}{27}\)\(=\dfrac{-2}{1.3}+\dfrac{-2}{3.5}+\dfrac{-2}{5.7}+...+\dfrac{-2}{19.21}+\dfrac{-2}{21.23}+\dfrac{-2}{23.25}+-\dfrac{2}{25.27}-\dfrac{1}{27}\)Đặt:
\(A=\dfrac{-2}{1.3}+\dfrac{-2}{3.5}+\dfrac{-2}{5.7}+...+\dfrac{-2}{19.21}+\dfrac{-2}{21.23}+\dfrac{-2}{23.25}+\dfrac{-2}{25.27}\)\(A=-1\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{25}-\dfrac{1}{27}\right)\)
\(A=-1\left(1-\dfrac{1}{27}\right)\)
\(A=-1.\dfrac{26}{27}=-\dfrac{26}{27}\)
Thay vào biểu thức ban đầu ta có:
Giá trị là:
\(-\dfrac{26}{27}-\dfrac{1}{27}=-\dfrac{27}{27}=-1\)