So Sánh
\(\dfrac{19}{40}\) và \(\dfrac{41}{80}\) \(\dfrac{19}{26}\) và \(\dfrac{21}{25}\)
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\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)
- Xét phân số trung gian: \(\frac{19}{25}\)
Ta có : \(\frac{19}{26}< \frac{19}{25}\)và \(\frac{19}{25}< \frac{21}{25}\)
\(\Rightarrow\frac{19}{26}< \frac{21}{25}\)
Hoặc có thể xét p/s trung gian là 21/26
phần còn lại trái dấu mình chưa ra
a) \(\dfrac{27}{35}>\dfrac{19}{35}>\dfrac{19}{41}\)
\(\Rightarrow\dfrac{27}{35}>\dfrac{19}{41}\)
b) \(\dfrac{120}{121}< \dfrac{120+1}{121+1}=\dfrac{121}{122}\)
\(\Rightarrow\dfrac{120}{121}< \dfrac{121}{122}\)
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.
\(\dfrac{5}{3}\cdot\dfrac{7}{25}+\dfrac{5}{3}\cdot\dfrac{21}{25}-\dfrac{5}{3}\cdot\dfrac{7}{25}\)
\(=\dfrac{5}{3}\cdot\left(\dfrac{7.}{25}+\dfrac{21}{25}-\dfrac{7}{25}\right)\)
\(=\dfrac{5}{3}\cdot\dfrac{21}{25}=\dfrac{7}{5}\)
b) \(250\%+19\dfrac{3}{11}\cdot\dfrac{7}{26}-6\dfrac{3}{11}\cdot\dfrac{7}{26}\)
\(=\dfrac{5}{2}+\dfrac{212}{11}\cdot\dfrac{7}{26}-\dfrac{69}{11}\cdot\dfrac{7}{26}\)
\(=\dfrac{7}{26}\cdot\left(\dfrac{212}{11}-\dfrac{69}{11}\right)+\dfrac{5}{2}\)
\(=\dfrac{7}{26}\cdot13+\dfrac{5}{2}\)
\(=\dfrac{7}{2}+\dfrac{5}{2}\)
\(=\dfrac{12}{2}=6\)
a) \(\dfrac{3}{5}+\dfrac{7}{25}=\dfrac{15}{25}+\dfrac{7}{25}=\dfrac{15+7}{25}=\dfrac{22}{25}\)
b) \(\dfrac{8}{11}-\dfrac{19}{33}=\dfrac{24}{33}-\dfrac{19}{33}=\dfrac{24-19}{33}=\dfrac{5}{33}\)
c) \(\dfrac{16}{21}\times\dfrac{3}{5}=\dfrac{16\times3}{21\times5}=\dfrac{48}{105}=\dfrac{16}{35}\)
d) \(\dfrac{14}{41}\div\dfrac{7}{9}=\dfrac{14}{41}\times\dfrac{9}{7}=\dfrac{14\times9}{41\times7}=\dfrac{126}{287}=\dfrac{18}{41}\)
\(\dfrac{26}{19}.\dfrac{4}{25}+\dfrac{7}{5}.\dfrac{3}{5}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{104}{475}+\dfrac{7}{5}.\dfrac{3}{5}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{104}{475}+\dfrac{21}{25}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{503}{475}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{503}{475}-\dfrac{28}{475}=\dfrac{475}{475}=1\)
\(\dfrac{26}{19}\cdot\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{3}{5}-\dfrac{4}{25}\cdot\dfrac{7}{19}=\dfrac{4}{25}\cdot\left(\dfrac{26}{19}-\dfrac{7}{19}\right)+\dfrac{7}{5}\cdot\dfrac{3}{5}=\dfrac{4}{25}\cdot1+\dfrac{21}{25}=\dfrac{4}{25}+\dfrac{21}{25}=\dfrac{25}{25}=1\)
\(-\dfrac{5}{7}\left(\dfrac{-19}{41}+-\dfrac{21}{41}+1\right)=-\dfrac{5}{7}.\dfrac{1}{41}=-\dfrac{5}{287}\)
\(=\dfrac{-5}{7}\left(\dfrac{19}{41}+\dfrac{21}{41}-1\right)=\dfrac{-5}{7}\cdot\dfrac{-1}{41}=\dfrac{5}{287}\)
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
a) Ta có :
\(\dfrac{19}{40}=\dfrac{38}{80}< \dfrac{41}{80}\)
b) Ta có :
\(\dfrac{19}{26}< \dfrac{21}{26}< \dfrac{21}{25}\)
41/80 lớn hơn 19/40
21/25 lớn hơn 19/ 26