a) \(\dfrac{-3}{100}>\dfrac{-50}{100}=-\dfrac{1}{2}\)

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

\(\Rightarrow\dfrac{-3}{100}>\dfrac{-2}{3}\)

b) \(\dfrac{-3}{5}=\dfrac{-9}{15}\)

\(\dfrac{-2}{3}=\dfrac{-10}{15}\)

Mà:  - 9 > -10 

\(\Rightarrow-\dfrac{9}{15}>\dfrac{-10}{15}\)

hay `-3/5>-2/3` 

c) \(\dfrac{-5}{4}< \dfrac{-2}{4}=-\dfrac{1}{2}\)

\(-\dfrac{3}{8}>\dfrac{-4}{8}=-\dfrac{1}{2}\)

\(\Rightarrow-\dfrac{5}{4}< \dfrac{-3}{8}\) 

d) \(-\dfrac{2}{3}=\dfrac{1}{3}-1\)

\(-\dfrac{3}{4}=\dfrac{1}{4}-1\)

Vì: `1/3>1/4` 

`=>1/3-1>1/4-1` 

Hay `-2/3>-3/4` 

a: \(\dfrac{-3}{100}=\dfrac{-3\cdot3}{100\cdot3}=\dfrac{-9}{300};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot100}{3\cdot100}=\dfrac{-200}{300}\)

mà -9>-200

nên \(\dfrac{-3}{100}>\dfrac{-2}{3}\)

b: \(\dfrac{-3}{5}=\dfrac{-3\cdot3}{5\cdot3}=\dfrac{-9}{15};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot5}{3\cdot5}=\dfrac{-10}{15}\)

mà -9>-10

nên \(\dfrac{-3}{5}>\dfrac{2}{-3}\)

c: \(\dfrac{-5}{4}=\dfrac{-5\cdot2}{4\cdot2}=\dfrac{-10}{8};\dfrac{-3}{8}=\dfrac{-3}{8}\)

mà -10<-3

nên \(-\dfrac{5}{4}< -\dfrac{3}{8}\)

d: \(\dfrac{-2}{3}=\dfrac{-2\cdot4}{3\cdot4}=\dfrac{-8}{12};\dfrac{3}{-4}=\dfrac{-3}{4}=\dfrac{-3\cdot3}{4\cdot3}=\dfrac{-9}{12}\)

mà -8>-9

nên \(-\dfrac{2}{3}>\dfrac{3}{-4}\)

e: \(\dfrac{267}{-268}=\dfrac{-267}{268}>-1;-1=\dfrac{-1343}{1343}>\dfrac{-1347}{1343}\)

Do đó: \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

f: \(\dfrac{2022\cdot2023-1}{2022\cdot2023}=1-\dfrac{1}{2022\cdot2023}\)

\(\dfrac{2023\cdot2024-1}{2023\cdot2024}=1-\dfrac{1}{2023\cdot2024}\)

Ta có: 2022<2024

=>\(2022\cdot2023< 2023\cdot2024\)

=>\(\dfrac{1}{2022\cdot2023}>\dfrac{1}{2023\cdot2024}\)

=>\(-\dfrac{1}{2022\cdot2023}< -\dfrac{1}{2023\cdot2024}\)

=>\(\dfrac{-1}{2022\cdot2023}+1< \dfrac{-1}{2023\cdot2024}+1\)

=>\(\dfrac{2022\cdot2023-1}{2022\cdot2023}< \dfrac{2023\cdot2024-1}{2023\cdot2024}\)

g: \(\dfrac{2022\cdot2023}{2022\cdot2023+1}=1-\dfrac{1}{2022\cdot2023+1}\)

\(\dfrac{2023\cdot2024}{2023\cdot2024+1}=1-\dfrac{1}{2023\cdot2024+1}\)

Vì \(2022\cdot2023+1< 2023\cdot2024+1\)

nên \(\dfrac{1}{2022\cdot2023+1}>\dfrac{1}{2023\cdot2024+1}\)

=>\(\dfrac{-1}{2022\cdot2023+1}< \dfrac{-1}{2023\cdot2024+1}\)

=>\(\dfrac{-1}{2022\cdot2023+1}+1< \dfrac{-1}{2023\cdot2024}+1\)

=>\(\dfrac{2022\cdot2023}{2022\cdot2023+1}< \dfrac{2023\cdot2024}{2023\cdot2024+1}\)

toán 7 hay 8 v

toán lớp 1 à

\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\\ =\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\\ =\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{5}{9}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{27}{45}+\dfrac{25}{45}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\dfrac{12}{12}+\dfrac{45}{45}+\dfrac{1}{35}\\ =1+1+\dfrac{1}{35}\\ =2+\dfrac{1}{35}\\ =\dfrac{70}{35}+\dfrac{1}{35}=\dfrac{71}{35}\)

\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{1}{35}\right)+\left(-\dfrac{7}{45}+\dfrac{5}{9}\right)\)

\(=\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{21}{35}+\dfrac{1}{35}\right)+\left(-\dfrac{7}{45}+\dfrac{25}{45}\right)\)

\(=1+\dfrac{22}{35}+\dfrac{18}{45}\)

\(=\dfrac{315}{315}+\dfrac{198}{315}+\dfrac{126}{315}\)

\(=\dfrac{71}{35}\)

$1,4$ giờ $\times4+3$ giờ $12$ phút $:6$

$=5,6$ giờ $+3,2$ giờ $:6$

$=5,6$ giờ $+\frac{8}{15}$ giờ

$=5$ giờ $36$ phút + $32$ phút

$=5$ giờ $68$ phút $=6$ giờ $8$ phút

$---$

$1\frac34$ giờ $+2$ giờ $18$ phút $\times 5-1,5$ giờ

$=1,75$ giờ $+2,3$ giờ $\times5-1,5$ giờ

$=(1,75$ giờ $-1,5$ giờ$)+11,5$ giờ

$=0,25$ giờ $+11,5$ giờ

$=11,75$ giờ $= 11$ giờ $45$ phút

a: 1,4 giờ x4+3h12p:6

=5,6 giờ+192p:6

=5,6 giờ+32p

=5h36p+32p=6h8p

b: Bạn ghi lại đề đi bạn

tại em non

Thuế VAT mà bác Minh phải trả khi mua chiếc điện thoại là:

\(10\%\cdot7990000=799000\left(đ\right)\) 

Số tiền mà bác Minh phải trả khi mua chiếc điện thoại là:

\(7990000+799000=8789000\left(đ\right)\) 

Trong sữa có \(3 , 6 \%\) bơ. Tính lượng sữa trong một chai, biết rằng lượng bơ trong chai sữa này là \(14 , 4\) g.

Số học sinh xếp loại trung bình của khối 4 trường đó là :

\(72:4=18\) ( học sinh )

Số học sinh khá và giỏi của khối 4 trường đó là :

\(72-18=54\) ( học sinh )

Đáp số : 54 học sinh

Số học sinh xếp loại trung bình của khối 4 là:

$72\times\frac14=18$ (học sinh)

Số học sinh được xếp loại khá và giỏi của khối 4 là:

$72-18=54$ (học sinh)

/s/: stops, arts, laughs, rocks

/iz/: quizzes, fixes, watches

/z/: friends, prefers

/ɪz/: quizzes, fixes, watches

/s/: stops, arts, laughs, rocks

/z/: friends, prefers

\(B=\left(1-\dfrac{1}{2^2}\right)\cdot\left(1-\dfrac{1}{3^2}\right)\cdot\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{2024^2}\right)\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot...\cdot\dfrac{2024^2-1}{2024^2}\)

Ta có CT: \(a^2-1=\left(a+1\right)\left(b+1\right)\)

\(B=\dfrac{\left(2+1\right)\left(2-1\right)}{2^2}\cdot\dfrac{\left(3+1\right)\left(3-1\right)}{3^2}\cdot\dfrac{\left(4+1\right)\left(4-1\right)}{4^2}...\cdot\dfrac{\left(2024+1\right)\left(2024-1\right)}{2024^2}\) 

\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{4\cdot2}{3^2}\cdot\dfrac{5\cdot3}{4^2}\cdot...\cdot\dfrac{2025\cdot2023}{2024^2}\)

\(=\dfrac{1\cdot2\cdot3^2\cdot...\cdot2023^2\cdot2024\cdot2025}{2^2\cdot3^2\cdot...\cdot2024^2}\)

\(=\dfrac{2025}{2\cdot2024}=\dfrac{2025}{4048}>\dfrac{2024}{4048}=\dfrac{1}{2}\)

Vậy: ...  

Ta có : 

\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{2024^2}\right)\)

\(=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}.....\dfrac{2024^2-1}{2024^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{2023.2025}{2024^2}\)

\(=\dfrac{1.2.3.....2023}{2.3.4.....2024}.\dfrac{3.4.5.....2025}{2.3.4.....2024}\)

\(=\dfrac{1}{2024}.\dfrac{2025}{2}=\dfrac{2025}{4048}>\dfrac{1}{2}\)

Vậy \(B>\dfrac{1}{2}\)