( 3 ¹⁹ . 4 + 5 . 3 ¹⁹ ) : 3 2¹

= [ 3 ¹⁹ . ( 4 + 5 ) ] : 3 ²¹

= ( 3 ¹⁹ . 9 ) : 3 ²¹

= ( 3 ¹⁹ . 3² ) : 3 ²¹

= 3 ²¹ : 3²¹

= 1

11111111

=11h39p

ra phép tính hẳn nha

=(2^2-1)*(2^2+1)*(2^4+1)*...*(2^64+1)
=(2^4-1)(2^2+1)*(2^4+1)*...*(2^64+1)
=(2^8-1)(2^8+1)*...*(2^64+1)
=...
=(2^64-1)(2^64+1)
=2^128-1

mk vua lm ra do!!!!!!

tks bạn nhưng mk làm ra lâu rồi sorry nha 

(mặc dù mk vẫn tk cho bạn)

\(=\)\(\left(2^2-1\right)\times\)\(\frac{2^4-1}{2^2-1}\)\(\times\frac{2^8-1}{2^4-1}\)\(\times\)...\(\times\)\(\frac{2^{128}-1}{2^{64}-1}\)

\(=\)\(2^{128}\)\(-1\)

\(19\cdot\left(-8\right)+19\cdot\left(-2\right)=19\cdot\left(-8+-2\right)=19\cdot-10=-190\)

``19.(-8)+19.(-2)`
`=19.[(-8)+(-2)]`
`=19.(-10)=-190`

19.(-8) + 19.(-2) = 19.(-8 + -2)

= 19.(-10)

= -190

HPT : \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{5}{36}\\\frac{4}{x}+\frac{3}{y}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{3}{x}+\frac{3}{y}=\frac{5}{12}\left(1\right)\\\frac{4}{x}+\frac{3}{y}=\frac{1}{2}\left(2\right)\end{cases}}\)

Từ (1) và (2), lấy vế trừ vế ta được :

\(\Leftrightarrow\left(\frac{4}{x}+\frac{3}{y}\right)-\left(\frac{3}{x}+\frac{3}{y}\right)=\frac{1}{2}-\frac{5}{12}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{12}\)

\(\Leftrightarrow\frac{1}{y}=\frac{5}{36}-\frac{1}{x}=\frac{5}{36}-\frac{1}{12}=\frac{1}{18}\)

\(\Leftrightarrow\hept{\begin{cases}x=12\\y=18\end{cases}}\)

1: \(\dfrac{-7}{4}\cdot\dfrac{2}{9}=\dfrac{-14}{36}=\dfrac{-7}{18}\)

2: \(=\dfrac{12}{13}\cdot\dfrac{26}{5}=\dfrac{24}{5}\)

3: \(=\dfrac{20}{11}\cdot\dfrac{55}{21}=\dfrac{100}{21}\)

4: \(=\dfrac{-40}{240}=\dfrac{-1}{6}\)

-4 nha bạn !