a) \(18+25\times4-4^3\)

\(=18+100-64\)

\(=118-64\)

\(=54\)

b) \(275-\left(49+125\div5^3\right)\)

\(=275-\left(49+125\div125\right)\)

\(=275-\left(49+1\right)\)

\(=275-49-1\)

\(=226-1\)

\(=225\)

c) \(2015+\left(8\times15-\left(18-8\right)^2\right)\)

\(=2015+\left(8\times15-\left(10\right)^2\right)\)

\(=2015+\left(8\times15-100\right)\)

\(=2015+\left(120-100\right)\)

\(=2015+20\)

\(=2035\)

Lời giải:

Vì $y^2\geq 0$ với mọi số tự nhiên $y$ nên:

$10(x-2019)^2=81-y^2\leq 81<90$

$(x-2019)^2<9$

$-3< x-2019< 3$

$2016< x< 2021$. Vì $x$ là số tự nhiên nên $x\in\left\{2017; 2018; 2019;2020\right\}$

Nếu $x=2017$ thì $y^2=41$ (loại vì $y$ tự nhiên)

Nếu $x=2018$ thì $y^2=71$ (loại vì $y$ tự nhiên)

Nếu $x=2019$ thì $y^2=81$ nên $y=9$

Nếu $x=2020$ thì $y^2=71$ (loại vì $y$ tự nhiên)

Vậy $x=2019; y=9$

Bài 2: 

a: \(\left(6x-39\right):7=3\)

\(\Leftrightarrow6x-39=21\)

hay x=10

a, 165:11.999 + 85.999 = (15.11):11.999 + 85.999 = [(15.11):11].999 + 85.999

= 15.(11:11).999 + 85.999 = 15.999 + 85.999 = (15+85).999 = 99900

b, 25.4 + 25:5 – 4.(30 – 5) – 5 = 25.4 + 25:5 – 4.25 – 5 = (25.4 – 4.25) + (25:5 – 5) = 0 + 0 = 0

Answer:

\(\dfrac{4}{-5}.2,7-0,8.6,3-80\%\)

\(=\dfrac{-4}{5}.\dfrac{27}{10}-\dfrac{4}{5}.\dfrac{63}{10}-\dfrac{4}{5}\)

\(=\dfrac{4}{5}.\dfrac{-27}{10}-\dfrac{4}{5}.\dfrac{63}{10}-\dfrac{4}{5}.1\)

\(=\dfrac{4}{5}.\left(\dfrac{-27}{10}-\dfrac{63}{10}-\dfrac{10}{10}\right)\)

\(=\dfrac{4}{5}.\dfrac{-27-63-10}{10}\)

\(=\dfrac{4}{5}.\dfrac{-100}{10}\)

\(=\dfrac{4}{5}.\left(-10\right)\)

\(=\dfrac{4}{1}.\dfrac{-2}{1}\)

\(=-8\)