-6,35 x 0,25 - 6,35 x 1,75 + 6,35

= - 6,35 x (0,25 + 1,75 - 1)

= -6,35 x (2 - 1)

= -6,35 x 1

= - 6,35

\(-6,35\cdot0,25-6,35\cdot1,75+6,35\)

\(=6,35\left(-0,25-1,75+1\right)\)

\(=6,35\cdot\left(-2+1\right)\)

\(=6,35\cdot\left(-1\right)=-6,35\)

\(\dfrac{x-1}{2023}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)

=>\(\dfrac{x-1}{2023}-\left(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

=>\(\dfrac{x-1}{2023}-2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{240}\right)=\dfrac{5}{8}\)

=>\(\dfrac{x-1}{2023}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

=>\(\dfrac{x-1}{2023}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

=>\(\dfrac{x-1}{2023}-2\cdot\dfrac{3}{16}=\dfrac{5}{8}\)

=>\(\dfrac{x-1}{2023}=\dfrac{5}{8}+\dfrac{3}{8}=\dfrac{8}{8}=1\)

=>x-1=2023

=>x=2024

-16 x (-0,125) x (-0,5)

= 2 x (-0,5)

= - 1

=-1 nha

A = 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + ... + 2018 - 2019 - 2020

Xét dãy số: 2; 3; 4;...; 2019; 2020

Dãy số trên có số số hạng là:

(2020 - 2) : 1 + 1 = 2019

Vì 2019 : 4 = 504 dư 3

Nhóm 4 số hạng liên tiếp của A vào nhau khi đó:

A = (2-3-4+5)+(6-7-8+9)+ ...+(2014 - 2015 - 2016+2017) + (2018 - 2019 - 2020)

A = 0 + 0 + ...+ 0 + (-1 - 2020)

A = - 2021

A = \(\frac13\) + \(\frac16\) + \(\frac{1}{10}\) + \(\frac{1}{15}\) + ... + \(\frac{1}{45}\)

A = \(\frac26\) + \(\frac{2}{12}\) + \(\frac{2}{20}\) + \(\frac{2}{30}\) + ... + \(\frac{2}{90}\)

A = 2.(\(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + ... + \(\frac{1}{9.10}\))

A = 2.(\(\frac12-\frac13\) + \(\frac13-\frac14\) +\(\frac14-\) \(\frac15\) + ... + \(\frac19\) - \(\frac{1}{10}\))

A = 2.(\(\frac12-\frac{1}{10}\))

A = 2.\(\frac25\)

A = \(\frac45\)

\(\dfrac{7}{12}-\dfrac{9}{20}=\dfrac{35}{60}-\dfrac{27}{60}=\dfrac{8}{60}=\dfrac{2}{15}\)

Giải:

Vì : 131 000 : 8 000 = 16 dư 3 000

Vậy Lan có thể mua nhiều nhất là 16 cái bút.

a: Trong cùng một nửa mặt phẳng bờ chứa tia Ox, ta có: \(\widehat{xOt}< \widehat{xOy}\left(50^0< 120^0\right)\)

nên tia Ot nằm giữa hai tia Ox và Oy

=>\(\widehat{xOt}+\widehat{yOt}=\widehat{xOy}\)

=>\(\widehat{yOt}=120^0-50^0=70^0\)

b: Ta có: \(\widehat{xOy}+\widehat{x'Oy}=180^0\)(hai góc kề bù)

=>\(\widehat{x'Oy}=180^0-120^0=60^0\)

Ta có: \(\widehat{xOt}+\widehat{x'Ot}=180^0\)(hai góc kề bù)

=>\(\widehat{x'Ot}=180^0-50^0=130^0\)

\(x>9,99\) nên \(x\in\left\{10,11,12,...\right\}\)

Mà x bé nhất nên \(x=10\)

10,00000000000..1

\(\dfrac{3}{5}+\dfrac{4}{7}\times\dfrac{14}{6}\)

\(=\dfrac{3}{5}+\dfrac{56}{42}\)

\(=\dfrac{3}{5}+\dfrac{4}{3}=\dfrac{9}{15}+\dfrac{20}{15}=\dfrac{29}{15}\)