a) `2/3` của 14 là:

\(\dfrac{2}{3}\times14=\dfrac{28}{3}\)

b) `5/4` của 451 là:

\(\dfrac{5}{4}\times451=\dfrac{2255}{4}\)

c) Đổi: \(1\dfrac{1}{3}=\dfrac{4}{3}\)

`4/3` của 60: 

\(\dfrac{4}{3}\times60=80\)

d) 23% của 50 là:

\(23\%\times50=11,5\)

16x+40=120

=>16x=120-40=80

=>\(x=\dfrac{80}{16}=5\)

\(16x+40=120\)

\(16x=120-40\)

\(16x=80\)

\(x=80:16\)

\(x=5\)

Vậy...

2x = 4048

=> x = 4048 : 2

=> x = 2024 

Vậy x = 2024 

x=1024

\(\dfrac{4}{15}< \dfrac{x}{30}< \dfrac{1}{3}\)

=>\(\dfrac{8}{30}< \dfrac{x}{30}< \dfrac{10}{30}\)

=>8<x<10

=>x=9

\(\dfrac{4}{15}< \dfrac{x}{30}< \dfrac{1}{3}\\ =>\dfrac{8}{30}< \dfrac{x}{30}< \dfrac{10}{30}\\ =>8< x< 10\)

Ta có:

\(\left|x-9\right|+\left|2-x\right|\ge\left|x-9+2-x\right|=\left|-7\right|=7\)

Dấu "=" xảy ra:

\(\left(x-9\right)\left(2-x\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-9\ge0\\2-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-9\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow2\le x\le9\)

\(\dfrac{x-1}{1}+\dfrac{x-1}{2}=\dfrac{x}{3}+\dfrac{x}{4}-\dfrac{7}{12}\\ =>x-1+\dfrac{x}{2}-\dfrac{1}{2}=\dfrac{x}{3}+\dfrac{x}{4}-\dfrac{7}{12}\\ =>\left(x+\dfrac{x}{2}\right)+\left(-1-\dfrac{1}{2}\right)=\left(\dfrac{x}{3}+\dfrac{x}{4}\right)-\dfrac{7}{12}\\ =>\dfrac{3}{2}x-\dfrac{3}{2}=\dfrac{7x}{12}-\dfrac{7}{12}\\ =>\dfrac{3}{2}x-\dfrac{7}{12}x=-\dfrac{7}{12}+\dfrac{3}{2}\\ =>\dfrac{11}{12}x=\dfrac{11}{12}=\\ =>x=\dfrac{11}{12}:\dfrac{11}{12}\\ =>x=1\)

\(\dfrac{3}{4}:1\dfrac{1}{14}+\dfrac{-2}{3}:\dfrac{4}{15}\\ =\dfrac{3}{4}:\dfrac{15}{14}+\dfrac{-2}{3}:\dfrac{4}{15}\\ =\dfrac{3}{4}\cdot\dfrac{14}{15}+\dfrac{-2}{3}\cdot\dfrac{15}{4}\\ =\dfrac{7}{10}+\dfrac{-5}{2}\\ =\dfrac{7}{10}+\dfrac{-25}{10}\\ =\dfrac{-18}{10}=-\dfrac{9}{5}\)

Cách 1: Chỉ ra tính chất đặc trưng 

\(B=\left\{x=2k;k\in N|k\le6\right\}\)

Cách 2: liệt kê

\(B=\left\{0;2;4;6;8;10;12\right\}\)

Cách 1: Liệt kê: 

`B =` {`0;2;4;6;8;10;12`}

Cách 2: Đặc trưng:

`B =` {`x` thuộc `N | x ⋮ 2`` và `x ≤13`}

\(\left(x-3\right)^2=\left(1-3x\right)^2\)

=>\(\left(3x-1\right)^2-\left(x-3\right)^2=0\)

=>\(\left(3x-1-x+3\right)\left(3x-1+x-3\right)=0\)

=>(2x+2)(4x-4)=0

=>\(\left[{}\begin{matrix}2x+2=0\\4x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

`(x - 1954) . 5 = 50`

`=> x - 1954 = 50 : 5`

`=> x - 1954 = 10`

`=> x = 10 + 1954`

`=> x = 1964`

Vậy `x = 1964`

`x - 152 : 2 = 46`

`=> x - 76 = 46`

`=> x = 46 + 76`

`=> x = 122`

Vậy `x = 122`

`70 - 5 (x - 3)=45`

`=> 5(x-3)=70-45`

`=> 5(x-3)=25`

`=> x-3=25 : 5`

`=> x - 3 = 5`

`=> x = 5+3`

`=> x = 8`

Vậy `x = 8`

`60-3(x-2)=51`

`=> 3(x - 2)=60-51`

`=> 3(x-2) = 9`

`=> x - 2 = 9:3`

`=> x -2 = 3`

`=> x = 3+2`

`=> x = 5`

Vậy `x = 5`

`10+2x=45.43`

`=> 10 +2x=1935`

`=> 2x = 1935 - 10`

`=> 2x=1925`

`=> x = 1925 : 2`

`=> x =` \(\dfrac{1925}{2}\)

Vậy `x =` \(\dfrac{1925}{2}\)

`4x - 20=25:23`

`=> 4x - 20 =` \(\dfrac{25}{23}\)

`=> 4x =`  \(\dfrac{25}{23}\) `+ 20`

`=> 4x =` \(\dfrac{485}{23}\)

`=> x =` \(\dfrac{485}{23}:4\)

`=> x =` \(\dfrac{485}{23}.\dfrac{1}{4}\)

`=>x=` \(\dfrac{485}{92}\)

Vậy `x =` \(\dfrac{485}{92}\)

c) ( x - 1954 ).5 = 50

 x - 1954 = 10

x = 1964

Vậy...

d) x - 152:2 = 46

x - 152 = 92

x = 244

Vậy...

e) 70 - 5 ( x- 3) = 45

65 ( x - 3 ) = 45

x - 3 = 9/13

x = 48/13

Vậy...

g) 60 - 3 ( x - 2 ) = 51

57 ( x - 2 ) = 51

x - 2 =17/19

x = 55/19

Vậy...

f) 10 + 2x = 45.43

10 + 2x = 1935

2x = 1925

x = 1925/2

Vậy...

h) 4x - 20 = 25:23

4x - 20 = 25/23

4x = 485/23

x = 485/92

Vậy...