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Bài 3:

a) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}=\dfrac{4}{3}-\dfrac{-4}{3}=\dfrac{8}{3}\)  

b) \(2\dfrac{1}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+0,4.5\right)\right]\) 

\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+\dfrac{2}{5}.5\right)\right]\) 

\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+2\right)\right]\)

\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\dfrac{8}{3}\right]\) 

\(=\dfrac{7}{3}-\dfrac{1}{3}.\dfrac{7}{6}\) 

\(=\dfrac{7}{3}-\dfrac{7}{18}\) 

\(=\dfrac{35}{18}\) 

c) \(\left(20+9\dfrac{1}{4}\right):2\dfrac{1}{4}=\left(20+\dfrac{37}{4}\right):\dfrac{9}{4}=\dfrac{117}{4}:\dfrac{9}{4}=13\) 

d) \(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\) 

\(=\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\) 

\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{32}{5}\) 

\(=10-\dfrac{32}{5}\) 

\(=\dfrac{18}{5}\) 

e) \(\dfrac{32}{15}:\left(-1\dfrac{1}{5}+1\dfrac{1}{3}\right)=\dfrac{32}{15}:\left(\dfrac{-6}{5}+\dfrac{4}{3}\right)=\dfrac{32}{15}:\dfrac{2}{15}=16\) 

g) \(0,2.\dfrac{15}{36}-\left(\dfrac{2}{5}+\dfrac{2}{3}\right):1\dfrac{1}{5}\) 

\(=\dfrac{1}{5}.\dfrac{5}{12}-\dfrac{16}{15}:\dfrac{6}{5}\) 

\(=\dfrac{1}{12}-\dfrac{8}{9}\) 

\(=\dfrac{-29}{36}\) 

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{8}{15}+0,25\right).\dfrac{24}{27}\) 

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{8}{9}\) 

\(=\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{8}{9}\) 

\(=\dfrac{7}{5}-\dfrac{94}{135}\) 

\(=\dfrac{19}{27}\) 

g) \(5:\left(4\dfrac{3}{4}-1\dfrac{25}{28}\right)-1\dfrac{3}{8}:\left(\dfrac{3}{8}+\dfrac{9}{20}\right)\) 

\(=5:\left(\dfrac{19}{4}-\dfrac{53}{28}\right)-\dfrac{11}{8}:\dfrac{33}{40}\) 

\(=5:\dfrac{20}{7}-\dfrac{5}{3}\) 

\(=\dfrac{7}{4}-\dfrac{5}{3}\) 

\(=\dfrac{1}{12}\)

\(a\)) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}\)

\(=\dfrac{4}{3}-\dfrac{7}{5}.\dfrac{-20}{21}\)

\(=\dfrac{4}{3}-\dfrac{-4}{3}\)

\(=\dfrac{8}{3}\)

\(=2\dfrac{2}{3}\)

bài đâu

Bài đou bạn :)?

lạc hậu

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bạn viết j thế

Bài 10:

a; Xét ΔEAB và ΔEGD có

\(\hat{EAB}=\hat{EGD}\) (hai góc so le trong, AB//GD)

\(\hat{AEB}=\hat{GED}\) (hai góc đối đỉnh)

Do đó: ΔEAB~ΔEGD

=>\(\frac{EA}{EG}=\frac{EB}{ED}\) (1)

Xét ΔEAD và ΔEKB có

\(\hat{EAD}=\hat{EKB}\) (hai góc so le trong, AD//KB)

\(\hat{AED}=\hat{KEB}\) (hai góc đối đỉnh)

Do đó: ΔEAD~ΔEKB

=>\(\frac{EA}{EK}=\frac{ED}{EB}\)

=>\(\frac{EK}{EA}=\frac{EB}{ED}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{EA}{EG}=\frac{EK}{EA}\)

=>\(EA^2=EK\cdot EG\)

b: \(\frac{1}{AK}+\frac{1}{AG}=\frac{AK+AG}{AK\cdot AG}=\frac{AE+EK+AE+EG}{\left(AE+EK\right)\left(AE+EG\right)}\)

\(=\frac{2\cdot AE+EK+EG}{AE^2+_{}AE\cdot\left(EK+EG\right)+EK\cdot EG}=\frac{2\cdot AE+EK+EG}{2\cdot AE^2+AE\cdot\left(EK+EG\right)}\)

\(=\frac{2\cdot AE+EK+EG}{AE\left(2\cdot AE+EK+EG\right)}=\frac{1}{AE}\)

đã nói kiểm tra ko giúp rồi mà:)

Bạn ko giúp thì thôi mik tự làm vâỵ🙂