Số lẻ nhỏ nhất từ 1 đến 104 là 1

Số lẻ lớn nhất từ 1 đến 104 là 103 

Khoảng cách giữa 2 số lẻ là: 2

Từ 1 đến 104 có số lượng số lẻ là:

(103 - 1) : 2 + 1 = 52 (số) 

lấy số cuối - số đầu: khoảng cách +1 là ra

\(B=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)

\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)

\(=\left(5-6-2\right)+\left(-\dfrac{3}{4}-\dfrac{7}{4}+\dfrac{5}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)\)

\(=-3+\dfrac{-5}{4}+\dfrac{-7}{5}=-3-1,25-1,4=-5,65\)

Olm chào em, em chưa hiểu chỗ nào em nhỉ?

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

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

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

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

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

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

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

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

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

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

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

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

$1,4$ 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

Lời giải:

Vận tốc xuôi dòng: $AB:3$ (km/h) 

Vận tốc ngược dòng: $AB:5$ (km/h) 

Vì vận tốc xuôi dòng hơn vận tốc ngược dòng 2 lần vận tốc tự nhiên của dòng nước nên vận tốc dòng nước là:

$(AB:3-AB:5):2=\frac{AB}{15}$ (km/h) 

Cụm bèo trôi xuôi dòng hết:

$AB: \frac{AB}{15}=15$ (giờ)

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

\(B=7-\left|4x-3\right|\) 

Ta có: \(\left|4x-3\right|\ge0\forall x\) 

\(\Rightarrow B=7-\left|4x-3\right|\le7-0=7\forall x\)

Dấu "=" xảy ra khi: \(4x-3=0\Leftrightarrow x=\dfrac{3}{4}\)

Vậy: ... 

\(4^{x+2}.3^x=16.12^5\\ \Rightarrow4^{x+2}.3^x=4^2.4^5.3^5\\ \Rightarrow4^{x+2}.3^x=4^7.3^5\\ \Rightarrow\dfrac{4^{x+2}}{4^7}.\dfrac{3^x}{3^5}=1\\ \Rightarrow4^{x-5}.3^{x-5}=1\\ \Rightarrow12^{x-5}=1\\ \Rightarrow x-5=0\\ \Rightarrow x=5\)

\(4^{x+2}\cdot3^x=16\cdot12^5\)

\(\Rightarrow4^{x+2}\cdot3^x=4^2\cdot4^5\cdot3^5\)

\(\Rightarrow4^{x+2}\cdot3^x=4^7\cdot3^5\)

\(\Rightarrow\left\{{}\begin{matrix}4^{x+2}=4^7\\3^x=3^5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+2=7\\x=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\)

\(\Rightarrow x=5\)

Vậy: ...