\(A=29\dfrac{1}{2}\cdot\dfrac{2}{3}+39\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{2}\cdot\dfrac{2}{3}+\dfrac{118}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{118}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{59}{2}+\dfrac{5}{6}\)

\(=59\cdot\left(\dfrac{1}{3}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=\dfrac{5}{6}\cdot\left(59+1\right)=\dfrac{5}{6}\cdot60=50\)

mọi người giúp em với ạ em đang cần gấp lắm .

Đặt S = (-1) x ( -1)²x ....x (-1)²⁰¹¹

Ta có : S = ( -1 ) x ( -1)²x ....x (-1) ²⁰¹¹

                   = (-1)  x 1 x....x 1

                  = ( -1 ) 

Vậy S = (-1) 

( Vì (-1)²= 1  theo quy ước của máy tính nen tớ làm như trên!! Học tốt !!)

(-1).(-1)².(-1)³...(-1)²⁰¹⁰.(-1)²⁰¹¹

=(-1).1.(-1)...1.(-1)  (có 1006 số -1 và 1005 số 1)

=[(-1).(-1).(-1)...(-1)].(1.1.1...1)

=1.1=1

\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)

\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)

\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)

\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)

\(=\frac{10-1}{\frac{99}{11}}\)

\(=\frac{9}{9}\)

\(=1\)

\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\)                                                     \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)

\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\)                                             \(=\)      \(\frac{21}{5}\)X  \(\frac{10}{11}+\frac{57}{11}\) 

\(=\frac{33}{5}-1\)                                                              \(=\frac{42}{11}+\frac{57}{11}\)

\(=\frac{28}{5}\)                                                                       \(=\frac{99}{11}=9\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}=\frac{1}{6}\)

a: \(A=\dfrac{2}{3}-4\cdot\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)

b: \(B=\dfrac{-2+5}{6}\cdot11-7=\dfrac{11}{2}-7=-\dfrac{3}{2}\)

c: \(=1+\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{27}{4}+5-1\)

\(=\dfrac{1}{3}+\dfrac{27}{4}+5=\dfrac{145}{12}\)

d: \(D=\dfrac{2}{3}\cdot\dfrac{2}{5}-\dfrac{1}{5}\cdot\dfrac{1}{6}=\dfrac{4}{15}-\dfrac{1}{30}=\dfrac{7}{30}\)

a. \(A=\dfrac{1}{2}\cdot\dfrac{2}{3}+\dfrac{5}{10}\) \(B=\dfrac{7}{8}-\dfrac{1}{4}\cdot\dfrac{3}{2}\)

\(=\dfrac{1}{3}+\dfrac{1}{2}\) \(=\dfrac{7}{8}-\dfrac{3}{8}\\ =\dfrac{1}{2}\)

\(\Rightarrow A>B\)

b. \(C=\dfrac{2}{3}\cdot\dfrac{4}{5}+\dfrac{8}{15}-\dfrac{1}{15}\)

\(=\dfrac{8}{15}+\dfrac{7}{15}\\ =\dfrac{15}{15}\\ =1\)

\(D=\dfrac{1}{2}\cdot\dfrac{5}{6}+\dfrac{2}{3}\cdot\dfrac{3}{4}\)

\(=\dfrac{5}{12}+\dfrac{6}{12}\\ =\dfrac{11}{12}\)

\(\Rightarrow C>D\)

( 1- \(\frac{1}{2}\))x( 1- \(\frac{1}{3}\))x( 1- \(\frac{1}{4}\))x( 1- \(\frac{1}{5}\))x( 1- \(\frac{1}{6}\)).

= \(\frac{1}{2}\)x \(\frac{2}{3}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\)x \(\frac{5}{6}\).

= \(\frac{1\times2\times3\times4\times5}{2\times3\times4\times5\times6}\).

= \(\frac{1}{6}\).

Ăn có ...chơi có. ...

a) \(\left(\frac{3}{5}.\frac{5}{3}\right).\frac{8}{27}\)

=  \(1.\frac{8}{27}\)

=\(\frac{8}{27}\)

b) \(\frac{7}{19}.\left(\frac{1}{3}.\frac{2}{3}\right)\)

= \(\frac{7}{19}.\frac{2}{9}\)

=\(\frac{14}{81}\)

Không biết có đúng không? Tick mình nha !