`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=>21x+14 <= 3x-12`

`<=> 18x <=-26`

`<=>x <= -13/9`

Vậy `x<=-13/9`.

\(\Leftrightarrow3x-6+13⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{3;1;15;-11\right\}\)

3x+7⋮ x+2

3(x+2)+1⋮x+2

1⋮x+2

x+2Ư(1)=

⇔3x−6+13⋮x−2⇔3x−6+13⋮x−2

hay 

`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=> 21x+14<=3x-12`

`<=>18x <= -26`

`<=> x <=-13/9`

\(x\left(x-1\right)-\left(x^2+x\right)-3x\left(x+1\right)+3\left(x^2+2\right)+5x\)

\(=x^2-x-x^2-x-3x^2-3x+3x^2+6+5x\)

= 6

Vậy bt trên ko phụ thuộc vào biến

a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)

\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)

\(\frac{3}{7}x=\frac{22}{35}\)

\(x=\frac{49}{35}=1,4\)

b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)

\(x:\frac{1}{2}=1,5-\frac{1}{4}\)

\(x:\frac{1}{2}=\frac{5}{4}\)

\(x=\frac{5}{4}.\frac{1}{2}\)

\(x=\frac{5}{8}\)

Vậy ..

a.\(\frac{3}{7}.x-\frac{1}{35}=\frac{3}{5}\)

\(\frac{3}{7}.x=\frac{3}{5}+\frac{1}{35}\)

\(\frac{3}{7}.x=\frac{11}{35}\)

\(x=\frac{22}{35}+\frac{3}{7}\)

\(x=\frac{37}{35}\)

a. \(1,5-x:\frac{1}{2}=\frac{1}{4}\)

                \(x:\frac{1}{2}=1,5-\frac{1}{4}\)

               \(x:\frac{1}{2}=\frac{5}{4}\)

              \(x=\frac{5}{4}.\frac{1}{2}\)

               \(x=\frac{5}{8}\)

Ta có: \(\left|3x+1\right|+\left|3x-5\right|=\left|3x+1\right|+\left|5-3x\right|\ge\left|3x+1+5-3x\right|=6\)(1)

\(\frac{12}{\left(y+3\right)^2+2}\le\frac{12}{2}=6\)(2)

\(\left(1\right);\left(2\right)\Rightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{1}{3}\le x\le\frac{5}{3}\\y=-3\end{cases}}\)

Thanks bn nha !! Nka

1: =>x^2+4x+3-x^2-2x=7

=>2x=4

hay x=2

Ta có: \(2x\left(x-1\right)-3x\left(x-5\right)=-x^2+7\)

\(\Leftrightarrow2x^2-2x-3x^2+15x+x^2-7=0\)

\(\Leftrightarrow13x-7=0\)

\(\Leftrightarrow13x=7\)

hay \(x=\frac{7}{13}\)

Vậy: \(x=\frac{7}{13}\)