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1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{19}{18}+\dfrac{5}{18}\)
\(=\dfrac{24}{18}\)
\(=\dfrac{4}{3}\)
2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\dfrac{1}{15}+\dfrac{7}{15}\)
\(=\dfrac{8}{15}\)
3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
a) Giải
So sánh từng số hạng của A với B, ta thấy:
\(\dfrac{19}{41}< \dfrac{21}{41};\dfrac{23}{53}< \dfrac{23}{49}\) và \(\dfrac{29}{61}< \dfrac{33}{65}\) (vì 29.65 < 33.61)
\(\Rightarrow\dfrac{19}{41}+\dfrac{23}{53}+\dfrac{29}{61}< \dfrac{21}{41}+\dfrac{23}{49}+\dfrac{33}{65}\)
\(\Rightarrow A< B\)
Vậy A < B
b) Giải
Ta có: \(C=\dfrac{19^{20}+5}{19^{20}-8}=\dfrac{19^{20}-8+13}{19^{20}-8}=1+\dfrac{13}{19^{20}-8}\)
\(D=\dfrac{19^{21}+6}{19^{21}-7}=\dfrac{19^{21}-7+13}{19^{21}-7}=1+\dfrac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\) và \(13>0\)
\(\Rightarrow\dfrac{13}{19^{20}-8}< \dfrac{13}{19^{21}-7}\)
\(\Rightarrow1+\dfrac{13}{19^{20}-8}< 1+\dfrac{13}{19^{21}-7}\)
\(\Rightarrow\) \(C< D\)
Vậy C < D.
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
a)\(\dfrac{19}{48}-\dfrac{3}{40}=\dfrac{95}{240}-\dfrac{18}{240}=\dfrac{77}{240}\)
b)\(\dfrac{1}{6}+\dfrac{7}{27}+\dfrac{5}{18}=\dfrac{9}{54}+\dfrac{14}{54}+\dfrac{15}{54}=\dfrac{38}{54}=\dfrac{19}{27}\)
\(E=1.2.3+2.3.4+3.4.5+...+18.19.20\)
\(4E=1.2.3.4+2.3.4.4+3.4.5.4+...+18.19.20.4\)
\(4E=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+18.19.20.\left(21-17\right)\)
\(4E=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+18.19.20.21-17.18.19.20\)
\(4E=18.19.20.21\)
\(4E=143640\)
\(E=\frac{143640}{4}\)
\(E=35910\)
Chúc bạn học tốt ~
\(G=2.4.6+4.6.8+6.8.10+...+18.20.22\) ( xem lại đề có nhầm dấu ko nha bn )
\(8G=2.4.6.8+4.6.8.8+6.8.10.8+...+18.20.22.8\)
\(8G=2.4.6.8+4.6.8.\left(10-2\right)+6.8.10.\left(12-4\right)+...+18.20.22\left(24-16\right)\)
\(8G=2.4.6.8+4.6.8.10-2.4.6.8+6.8.10.12-4.6.8.10+...+18.20.22.24-16.18.20.22\)
\(8G=18.20.22.24\)
\(8G=190080\)
\(G=\frac{190080}{8}\)
\(G=23760\)
Chúc bạn học tốt ~
Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
Ta có:
\(A=\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
\(B=\frac{19^{21}+6}{19^{21}-7}=\frac{19^{21}-7+13}{19^{21}-7}=1+\frac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\Rightarrow\frac{13}{19^{20}-8}>\frac{13}{19^{21}-7}\)
\(\Rightarrow A>B\)
maivananh
A<B vì các số mũ đều giống nhau nên sẽ s2 phần số nguyên vậy nên nếu s2 5 với 6 thì 5 bé hơn nên A<B
giải chi tiết