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\(A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(A=\dfrac{\dfrac{21}{44}+\dfrac{3}{13}}{\dfrac{20}{77}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{6}+\dfrac{1}{4}}{\dfrac{5}{12}+\dfrac{5}{8}}\)
\(A=\dfrac{\dfrac{645}{1001}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\)
\(\Leftrightarrow A=\left(\dfrac{645}{1001}:\dfrac{645}{1001}\right)+\left(\dfrac{5}{12}:\dfrac{25}{24}\right)\)
\(A=1+\dfrac{2}{5}\)
\(A=\dfrac{7}{5}.\)
Vậy \(A=\dfrac{7}{5}.\)
có gì đó sai sai ở chỗ 5/7 thì phải . mik nghĩ đó 5/4 mới đúng chứ .
Lời giải:
Ta có:
\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\) \(=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
help
Sửa đề:
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(A=\dfrac{3}{5}.\dfrac{\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}}{\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}\right)}\)
\(A=\dfrac{3}{5}+\dfrac{1}{\dfrac{5}{2}}=\dfrac{3}{5}+\dfrac{2}{5}=1\)
Chúc bạn hcọ tốt!!!
\(\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}\)+\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{3}\right)}\)
=\(\frac{3}{5}\)+\(\frac{1}{\frac{5}{2}}\)
=\(\frac{3}{5}\)+\(\frac{2}{5}\)
=1 !!!
a,
Đặt A = \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\right)\)
\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{95}+...+\dfrac{1}{3}-1\right)\)
2A= \(2.\left(\dfrac{1}{99}-1\right)\)
\(\Rightarrow\) A = \(\dfrac{1}{99}-1=\dfrac{-98}{99}\)
b, \(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
= \(\dfrac{3.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}{5.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)
= \(\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{5}{5}=1\)
Chúc bn hc tốt <3
*Trả lời :
a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
=\(\dfrac{3}{4}.\left(-8\right)\)
= \(-6\)
b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
=\(10:\left(-\dfrac{5}{7}\right)\)
=\(-14\)
c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )
=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)
=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)
=\(\left(-1+1\right):\dfrac{7}{11}\)
\(=0:\dfrac{7}{11}\)
=0.
d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
\(a,13\dfrac{3}{5}-\left(8\dfrac{3}{5}-4\dfrac{3}{4}\right)\)
\(=\dfrac{68}{5}-\dfrac{43}{5}+\dfrac{19}{4}\)
\(=5+\dfrac{19}{4}\)
\(=\dfrac{20}{4}+\dfrac{19}{4}=\dfrac{39}{4}\)
\(------\)
\(b,11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)
\(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}\)
\(=\left(\dfrac{45}{4}-\dfrac{21}{4}\right)-\dfrac{19}{7}\)
\(=6-\dfrac{19}{7}\)
\(=\dfrac{42}{7}-\dfrac{19}{7}=\dfrac{23}{7}\)
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{2}{4}+\dfrac{2}{8}-\dfrac{2}{6}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{6}\right)}{5\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{7}\right)}=\dfrac{3}{5}+\dfrac{2}{5}=1\)