a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\)

<=> \(x=\dfrac{7}{12}-\dfrac{5}{24}=\dfrac{14}{24}-\dfrac{5}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\)

<=> \(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)

c) bn ghi rõ đề chút

Quá giỏi   ❤

\(\left|x+1\right|+\left|x-5\right|=3x+1\left(đk:x\ge-\dfrac{1}{3}\right)\)

\(\Leftrightarrow x+1+\left|x-5\right|=3x+1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x\left(x\ge5\right)\\x-5=-2x\left(-\dfrac{1}{3}\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-x-1+5-x=3x+1\left(x< -1\right)\\x+1+5-x=3x+1\left(-1\le x< 5\right)\\x+1+x-5=3x+1\left(x\ge5\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(ktm\right)\\x=\dfrac{5}{3}\left(tm\right)\\x=-5\left(ktm\right)\end{matrix}\right.\Rightarrow x=\dfrac{5}{3}\)

Lời giải:

$\frac{3}{8}+(x-\frac{5}{24}):\frac{1}{2}=1$

$(x-\frac{5}{24}):\frac{1}{2}=1-\frac{3}{8}=\frac{5}{8}$

$x-\frac{5}{24}=\frac{5}{8}\times \frac{1}{2}=\frac{5}{16}$

$x=\frac{5}{16}+\frac{5}{24}=\frac{25}{48}$

Câu a:

x - 1: 4/3 + 9/16 = - 5^2/4^2

x - 1 x 3/4 + 9/16 = - 25/16

x - 3/4 + 9/16 = - 25/16

x - 3/4 = -25/16 - 9/16

x - 3/4 = -17/8

x = - 17/8 + 3/4

x = -11/8

Vậy x = -11/8

Câu b:

5/2x + 2 - 3 1/2 = 1/6

5/2x + 2 - 7/2 = 1/6

5/2x = 1/6 + 7/2 - 2

5/2x = 1/6 + 21/6 - 12/6

5/2x = 5/3

x = 5/3 : 5/2

x = 2/3

\(\frac{4}{5}.x-x-\frac{3}{2}.x+\frac{5}{6}=\frac{1}{2}-\frac{4}{3}\)đề nảy hả

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

\(-\frac{1}{5}x-\frac{3}{2}x=-\frac{5}{6}-\frac{5}{6}\) 

\(-\frac{17}{10}x=-\frac{10}{6}\) 

\(x=\frac{50}{51}\)         

|x+3/5|-1/2=1/2

|x+3/5|=1/2+1/2

|x+3/5|=1

=>x+3/5=1

x=1-3/5

x=2/5

hay

x+3/5=-1

x=(-1)-3/5

x=-8/5

Vậy x=2/5 hay-8/5

1)vì |x-4/5|=3/4 =>x-4/5=3/4 hay x-4/5=-3/4 x=3/4+4/5 x= -3/4+4/5 x=31/20 x=1/20 vậy x=31/20 hay 1/20

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

\(=3x^2\cdot x^2-3x^2\cdot2x+3x^2\cdot5\)

\(=3x^4-6x^3+15x^2\)

b: (x+1)(2x-3)

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

\(=2x^2-x-3\)

c: \(\left(2x^3-3x^2+x+15\right):\left(2x+3\right)\)

\(=\left(2x^3+3x^2-6x^2-9x+10x+15\right):\left(2x+3\right)\)

\(=\frac{x^2\left(2x+3\right)-3x\left(2x+3\right)+5\left(2x+3\right)}{2x+3}=x^2-3x+5\)