Đổi: 6 tấn = 6000 kg

a, Tỉ số của gạo thu được so với lượng thóc ban đầu là:

\(\dfrac{3990}{6000}.100\%=66,5\%\)

a, \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)

=  \(\dfrac{3}{12}\) - \(\dfrac{2}{12}\) + \(\dfrac{1}{12}\)

=  \(\dfrac{2}{12}\)

=  \(\dfrac{1}{6}\)

b, \(\dfrac{3}{8}\) - ( \(\dfrac{3}{4}\) - \(\dfrac{1}{2}\))

= \(\dfrac{6}{16}\) - \(\dfrac{12}{16}\) + \(\dfrac{8}{16}\)

= \(\dfrac{2}{16}\)

= \(\dfrac{1}{8}\)

a) 1/4 - 1/6 = 1/12

= 3/12 - 2/12 + 1/12

=1/12 + 1/12

= 2/12

= 1/ 6

B) 3/8 - ( 3/4 - 1/2)

= 6/16 - (12 / 16 -8/16)

= 6/16 - 4 /16

= 2/16

=1/8

A = \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\) + \(\dfrac{3}{25}\) +..............+ \(\dfrac{3}{(3n)^2}\)

A = ( \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\)+ \(\dfrac{3}{25}\)) +.....+ \(\dfrac{3}{(3n)^2}\)

A = 3. ( \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\))+............+ \(\dfrac{3}{(3n)^2}\)

A = 3.( \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + \(\dfrac{1}{5.5}\)) +............+ \(\dfrac{3}{(3n)^2}\)

Vì \(\dfrac{1}{2}\) > \(\dfrac{1}{3}\) > \(\dfrac{1}{4}\) > \(\dfrac{1}{5}\)Ta có : \(\dfrac{1}{2.2}>\dfrac{1}{2.3}>\dfrac{1}{3.3}>\dfrac{1}{3.4}>\dfrac{1}{4.4}>\dfrac{1}{4.5}>\dfrac{1}{5.5}>\dfrac{1}{5.6}\)

A > 3. ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)) + ............+ \(\dfrac{1}{(3n)^2}\)

A > 3. ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)) +.....+ \(\dfrac{1}{(3n)^2}\)

A > 3.( \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) +..............+ \(\dfrac{1}{(3n)^2}\)

A > 3. \(\dfrac{1}{3}\) +...............+ \(\dfrac{1}{(3n)^2}\)

A > 1 +..........+ \(\dfrac{1}{9n^2}\) > 1 

A > 1 

\(S< \dfrac{1}{10x11}+\dfrac{1}{11x12}+...+\dfrac{1}{\left(n-1\right)n}\)

\(S< \dfrac{1}{10}-\dfrac{1}{n}< \dfrac{1}{10}\)

<

\(5⋮\left(2x-1\right)\)

\(\Rightarrow\left(2x-1\right)\in U\left(5\right)\in\left(\pm1;\pm5\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\text{=}-1\\2x-1\text{=}1\\2x-1\text{=}-5\\2x-1\text{ }\text{=}5\end{matrix}\right.\Leftrightarrow x\text{=}\left\{{}\begin{matrix}0\\1\\-2\\3\end{matrix}\right.\)

\(5⋮\left(2x-1\right)\)

=> 2x-1 thuộc ước của 5

mà \(Ư\left(5\right)\in\left\{-1;1;-5;5\right\}\)

ta có bảng sau

vậy \(x\in\left\{0;1;-2;3\right\}\)

\(x^2.x=\left(-9\right).3\)

\(\Rightarrow x^3=-27\)

\(\Rightarrow x^3=\left(-3\right)^3\)

\(\Rightarrow x=-3\)

Vậy x = -3.

$x^2.x=\left(-9\right).3$

$\Rightarrow x^3=-27$

$\Rightarrow x^3={\left(-3\right)}^3$

$\Rightarrow x=-3$

Vậy x = -3.

Vẽ được số đường thẳng đi qua các cặp điểm đó là:

\(\dfrac{4.3}{2}=6\) đường thẳng

6 đường thẳng đó là: AB, AC, AD, CB, BD, DC.

Đặt A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}...+\dfrac{1}{2^x}\) suy ra 2A= \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{x-1}}\) 

2A-A=2= \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{x-1}}\)-\(1-\dfrac{1}{2}-\dfrac{1}{2^2}...-\dfrac{1}{2^x}\)

A= \(2-\dfrac{1}{2^x}\)

Khi đó: \(\dfrac{1}{1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^x}}=\dfrac{1}{2-\dfrac{1}{2^x}}=\dfrac{2^x}{127}\) suy ra: 127=\(2^{x+1}-1\)=>127+1=128=\(2^7\)=\(2^{x+1}\)=>x+1=7=>x=6

Vậy x=6