ta có 

\(-\frac{3^7\cdot2^8}{6^7}=-\frac{3^7\cdot2^7\cdot2}{6^7}=-\frac{6^7\cdot2}{6^7}=-2\)2

\(B=-\frac{3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}=-\frac{3\cdot7^3\cdot7+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}=-\frac{7^3\cdot\left(3\cdot7+1\right)}{7^3\cdot\left(7^2\cdot6-2\right)}=-\frac{5}{73}\)

a) \(\dfrac{\left(-3\right)^7\cdot2^8}{6^7}\)

\(=\dfrac{-1\cdot3^7\cdot2^8}{\left(2\cdot3\right)^7}=\dfrac{-1\cdot3^7\cdot2^7\cdot2}{2^7\cdot3^7}=-1\cdot2=-2\)

b) \(\dfrac{-3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}\)

\(=\dfrac{-3\cdot7\cdot7^3+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}\)

\(=\dfrac{7^3\left(-3\cdot7+1\right)}{7^3\left(7^2\cdot6-2\right)}=\dfrac{-3\cdot7+1}{7^2\cdot6-2}\)

\(=\dfrac{-21+1}{294-2}=\dfrac{-20}{290}=\dfrac{-2}{29}\)

b) \(\dfrac{5^3\cdot3^5}{5^3\cdot0,5+125\cdot2\cdot5}\)

\(=\dfrac{5^3\cdot3^5}{5^3\cdot0,5+5^3\cdot2\cdot5}=\dfrac{5^3\cdot3^5}{5^3\left(0,5+2\cdot5\right)}\)

\(=\dfrac{3^5}{0,5+2\cdot5}=\dfrac{243}{10,5}=\dfrac{162}{7}\)

ta có : 6 + x = 7\(\frac{1}{2}\) + 2 - x 

<=> x + x = 7\(\frac{1}{2}\) + 2 - 6

<=> 2x = 15/2 + 4/2 - 12/2 

<=> 2x = 7/2

=> x = 7/2 : 2 

=> x = 7/4 

1024 : 2⁵+140 : ( 38 + 2⁵ ) + 7⁴ + 7²

= 1024 : 2⁵+140 : ( 38 + 32 ) + 7⁴ + 7²

⁼ 1024 : 2⁵+140 : 70 + 7⁴ + 7²

⁼ 1024 : 32 +140 : 70 + 2401 + 49

= 32  + 2 + 2401 + 49

= 34 + 2450

= 2484

\(A=-\left|x-7\right|+2\le2\\ A_{max}=2\Leftrightarrow x-7=0\Leftrightarrow x=7\\ B=-5-\left|2x+3\right|\le-5\\ A_{max}=-5\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\)

Chọn B

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