Tỉ số giữa 3x và x-2 là 1,5 nên ta có:

\(\dfrac{3x}{x-2}=1,5=\dfrac{3}{2}\)

=>\(\dfrac{x}{x-2}=\dfrac{1}{2}\)

=>2x=x-2

=>2x-x=-2

=>x=-2

Tìm \(x\) biết:

\(\frac{3x}{x-2}\) = 1,5

3\(x\) = (\(x\) - 2) x 1,5

3\(x\) = 1,5\(x\) - 3

3\(x\) - 1,5\(x\) = - 3

1,5\(x\) = -3

\(x=-3:1,5\)

\(x\) = - 2

Vậy \(x=-2\)

\(\frac59=\frac{20}{36}\); \(\frac16=\frac{6}{36}\); \(\frac{5}{12}\) = \(\frac{15}{36}\); \(\frac34=\frac{27}{36}\); Vì:

\(\frac{6}{36}<\frac{15}{36}<\frac{20}{36}<\frac{27}{36}\)

Vậy môn học được các bạn 6a yêu thích nhất là: môn hoạt động trải nghiệm.

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

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