\(\dfrac{5}{4x}-\dfrac{3}{2}=\dfrac{3}{2}+\dfrac{x}{2}\\ \Leftrightarrow5-6x=6x+2x^2\\ \Leftrightarrow2x^2+12x-5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-6+\sqrt{46}}{2}\\x=\dfrac{-6-\sqrt{46}}{2}\end{matrix}\right.\)

Với  x ≥ 0 , x ≠ 25  Ta có:  A = B . x − 4

⇔ x + 2 x − 5 = 1 x − 5 . x − 4 ⇔ x + 2 = x − 4 ( * )

Nếu  x ≥ 4 ,   x ≠ 25  thì (*) trở thành : x + 2 = x − 4

⇔ x − x − 6 = 0 ⇔ x − 3 x + 2 = 0

Do  x + 2 > 0  nên  x = 3 ⇔ x =   9  (thỏa mãn)

Nếu  0 ≤ x < 4  thì (*) trở thành : x + 2 = 4 − x

⇔ x + x − 2 = 0 ⇔ x − 1 x + 2 = 0

Do  x + 2 > 0  nên  x = 1 ⇔ x = 1  (thỏa mãn)

Vậy có hai giá trị x=1 và x= 9 thỏa mãn yêu cầu bài toán

\(a)\) \(2x-5=21\)

\(\Leftrightarrow\) \(2x=21+5\)

\(\Leftrightarrow\) \(2x=26\)

\(\Leftrightarrow\) \(x=26:2\)

\(\Leftrightarrow\) \(=13\)

\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)

\(\Leftrightarrow\) \(x=\frac{1}{3}\)

a)\(\frac{2}{3}\cdot x+\frac{1}{4}=\frac{5}{6}\)

\(\frac{2}{3}\cdot x=\frac{5}{6}-\frac{1}{4}=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac{2}{3}=\frac{7}{8}\)

b) \(\frac{7}{12}-\frac{5}{6}\cdot x=\frac{1}{4}:\frac{2}{3}\)

\(\frac{7}{12}-\frac{5}{6}\cdot x=\frac{3}{8}\)

\(\frac{5}{6}\cdot x=\frac{7}{12}-\frac{3}{8}=\frac{5}{24}\)

\(x=\frac{5}{24}:\frac{5}{6}=\frac{1}{4}\)

c) \(2\frac{1}{3}-\left(x+1\right)=\frac{5}{9}\)

\(\frac{7}{3}-\left(x+1\right)=\frac{5}{9}\)

\(x+1=\frac{7}{3}-\frac{5}{9}=\frac{16}{9}\)

\(x=\frac{16}{9}-1=\frac{7}{9}\)

d) \(\frac{2\cdot x+1}{15}=\frac{3}{5}\)

\(\left(2\cdot x+1\right):15=\frac{3}{5}\)

\(2\cdot x+1=\frac{3}{5}\cdot15=9\)

\(2\cdot x=9-1=8\)

\(x=8:2=4\)

a, 2/3 . x +1/4=5/6

      2/3 . x=5/6-1/4

      2/3 . x=10/12 -3/12

       2/3 . x=7/12

       x= 7/12 : 2/3

      x=7/8 

Vậy x=7/8

720:[41-(2×x-5)]=120

[41-(2×x-5)]=720:120

[41-(2×x-5)]=6

(2×x-5)=41-6

(2×x-5)=35

2×x=35+5

2x=40

x=40:2

vậy x=20

a) \(720:\left[41-\left(2x-5\right)\right]=120\)

\(\Rightarrow\left[41-\left(2.x-5\right)\right]=6\)

\(\Rightarrow2x-5=35\Rightarrow2x=40\Rightarrow x=20\)

b) \(4,8.5-x+100=250\)

\(\Rightarrow4,8.5-x=150\Rightarrow24-x=150\Rightarrow x=-126\)

c)+)  \(\frac{3}{5}+\frac{2}{5}:x=1\Rightarrow\frac{2}{5}:x=\frac{2}{5}\Rightarrow x=1\)

+) \(\frac{3}{5}+\frac{2}{5}:x=\frac{1}{4}\Rightarrow\frac{2}{5}:x=\frac{-7}{20}\Rightarrow x=\frac{-8}{7}\)

Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu hok tốt ^^

Với  x ≥ 0 , x ≠ 25  thì  B = 3 x + 5 + 20 − 2 x x − 15 = 3 x + 5 + 20 − 2 x x + 5 x − 5

= 3 x − 5 + 20 − 2 x x + 5 x − 5 = 3 x − 15 + 20 − 2 x x + 5 x − 5 = x + 5 x + 5 x − 5 = 1 x − 5

 (điều phải chứng minh)

`1/2 xx x +3/2 =7`

`=> 1/2 xx x = 7-3/2`

`=> 1/2 xx x = 14/2 -3/2`

`=> 1/2 xx x = 11/2`

`=> x= 11/2 :1/2`

`=> x=11/2 xx2`

`=> x= 22/2`

`=>x=11`

Vậy `x=11`

__

`3/2 xx x -2/7 xx(x-7/2)=18`

`=> 3/2 xx x -2/7x + 1=18`

`=> (3/2 -2/7 )x+ 1 =18`

`=> 17/14 x=18-1`

`=> 17/14x=17`

`=>x=17:17/14`

`=> x=17 xx 14/17`

`=>x=14`

a) \(\dfrac{1}{2}\times x+\dfrac{3}{2}=7\)

\(\dfrac{1}{2}\times x=7-\dfrac{3}{2}\)

\(\dfrac{1}{2}\times x=\dfrac{11}{2}\)

\(x=\dfrac{11}{2}\div\dfrac{1}{2}\\ x=\dfrac{11}{2}\times2\\ x=\dfrac{22}{2}=11\)

b) \(\dfrac{3}{2}\times x-\dfrac{2}{7}\times\left(x-\dfrac{7}{2}\right)=18\)

\(\dfrac{3}{2}\times x-\left(\dfrac{2}{7}x-1\right)=18\)

\(\dfrac{3}{2}\times x-\dfrac{2}{7}x+1=18\)

\(\dfrac{3}{2}x-\dfrac{2}{7}x+1=18\)

\(\dfrac{17}{14}x+1=18\)

\(\dfrac{17}{14}x=18-1\)

\(\dfrac{17}{14}x=17\)

\(x=17\div\dfrac{17}{14}\)

\(x=17\times\dfrac{14}{17}\)

\(x=14\)