\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14

2 (x-1) - 5 (x+2) = -10

2x-2 - 5x+10 = -10

2x-5x-2+10=-10

2x-5x=-10-10+2

-3x=-18

x=6

Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)

đkxđ:xx>3 

\(\left|5-2x\right|=x-4\) 

=>TH1:

\(5-2x=x-4\)

-x-2x=-5-4

-3x=-9

x=3(loại)

TH2:

5-2x=-x+4

x-2x=-5+4

-x=-1

x=1(loại)

vậy ko tìm đc x thỏa mãn đề bài

\(\left|5-2x\right|-3=x-7\)

\(\left|5-2x\right|=x-7+3\)

\(\left|5-2x\right|=x-4\)

Đk: \(x-4\ge0\)\(\Rightarrow x\ge4\)

Ta có: \(\left|5-2x\right|=x-4\)

\(\Rightarrow\orbr{\begin{cases}5-2x=x-4\\5-2x=-x+4\end{cases}\Rightarrow}\orbr{\begin{cases}-2x-x=-4-5\\-2x+x=4-5\end{cases}\Rightarrow}\orbr{\begin{cases}3x=9\\-x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)( cả 2 trường hợp x ko thỏa mãn )

Vậy \(x\in\varnothing\)

Nhầm :(

-2x + x = 4 - 5

<=> -x = -1

<=> x = 1 ( ko thỏa mãn )

\(3x-7+2\left(5-2x\right)+5x=19\)

\(\Leftrightarrow3x-7+10-4x+5x=19\)

\(\Leftrightarrow3x-4x+5x=19-10+7\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

Ta có \(\frac{x}{3}+\frac{x}{4}=7\)

\(\Leftrightarrow\frac{4x}{12}+\frac{3x}{12}=7\)

\(\Leftrightarrow\frac{7x}{12}=7\)

\(\Leftrightarrow7x=7.12\)

<=> x = 12

Vậy x = 12

Học tốt

x/3+x/4= 7

TRL:

12/3+12/4 = 7

# HỌC TỐT, NẾU ĐÚNG NHỚ K CHO MK#

x/3 + x/4 = 4*x/12 + 3*x/12

= 7*x/12 

=> 7*x = 7*12

=> x=12

Vậy x=12

(15/4 x 2/3) : 5=\(\dfrac{5}{2}\):5=1/2

\(\left(\dfrac{15}{4}.\dfrac{2}{1}\right):5=\dfrac{15}{2}:5=\dfrac{15}{2}.\dfrac{1}{5}=\dfrac{3}{2}\)

1/2 nhé

nhấn lộn phím