...

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

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

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

a)

`(2x-1)(x+2/3)=0`

\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b)

\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)

\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)

\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)

\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)

\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)

sai rồi , x không thể có 2 giá trị

`3/7 x -2/3 x =10/21`

`=> (3/7-2/3)x=10/21`

`=> ( 9/21 - 14/21)x=10/21`

`=>-5/21 x=10/21`

`=> x=10/21 : (-5/21)`

`=> x=10/21 xx (-21/5)`

`=>x=-2`

(3/7 - 2/3).x = 10/21

-5/21.x = 10/21

x = 10/21 : (-5/21) 

x = -2

vậy x = -2

Dễ

Dễ mà ko biét làm

Gọi chiều dài là x, rộng là y  ( x > y > 0)

Theo bài ra ta có : \(\dfrac{x}{\dfrac{1}{5}}\) = \(\dfrac{y}{\dfrac{1}{8}}\) ⇒ 5x = 8y ⇒ \(\dfrac{x}{8}\) = \(\dfrac{y}{5}\)

⇒ \(\dfrac{x^2}{64}\)= \(\dfrac{x.y}{8.5}\)= \(\dfrac{360}{40}\) = 9⇒ \(\) \(x\)² = 64.9 = 24² ⇒ x = 24; x= -24 (loại)

y = 360 : 24 = 15

Vậy chiều dài là 24m, rộng 15m 

`(x-1)/(x+5) = 6/7` ĐKXĐ : `x-5≠0<=>x≠5`

`<=> (7(x-1))/(7(x+5)) = (6(x+5))/(7(x+5))`

`=> 7(x-1)=6(x+5)`

`<=>7x-7=6x+30`

`<=>7x-6x=30+7`

`<=>x= 37 (TM)`

\(6\left(x+5\right)=7\left(x-1\right)\)

\(6x+30=7x-7\)

\(-x=-37\)

\(x=37\)

\(2x+3y-14=186\)

\(\Rightarrow2x+3y=186+14=200\)

Từ \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{18}\) suy ra \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{18}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{18}=\dfrac{2x+3y}{30+60}=\dfrac{20}{9}\)

\(\Rightarrow x=15\cdot\dfrac{20}{9}=\dfrac{100}{3}\)

\(\Rightarrow y=20\cdot\dfrac{20}{9}=\dfrac{400}{9}\)

\(z=18\cdot\dfrac{20}{9}=40\)

2x + 3y - 14 = 186 => 2x + 3y = 186 + 14 = 200

\(\dfrac{x}{15}\) = \(\dfrac{y}{20}\) = \(\dfrac{z}{18}\) ⇒ \(\dfrac{2x}{30}\) = \(\dfrac{3y}{60}\) = \(\dfrac{2x+3y}{30+60}\) = \(\dfrac{200}{90}\)  = \(\dfrac{20}{9}\) 

=> x = \(\dfrac{20}{9}\) x 30 : 2 = \(\dfrac{100}{3}\);     y =   \(\dfrac{20}{9}\) x 60 : 3 = \(\dfrac{400}{9}\)

z = \(\dfrac{100}{3}\) : 15 x 18 = 40 

Vậy (x, y, z) =( \(\dfrac{100}{3}\); \(\dfrac{400}{9}\); 40)