\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)

x=-36

x=11

x=100

x=14

x=-36

x=11

x=100

x=14

a) Ta có : ( x + 1 ).( 3 - x ) > 0

Th1 : \(\hept{\begin{cases}x+1>0\\3-x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x>3\end{cases}\Rightarrow}x>3}\)

Th2 : \(\hept{\begin{cases}x+1< 0\\3-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x< 3\end{cases}\Rightarrow}x< -1}\)

sao ko ai làm giúp mk vậy

Giúp mình nhé các bạn mình đang cần gấp lắm

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

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

=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

c , d giải nốt:

c) 2x + 3/6 = 7x - 13/15

=> (2x + 3) . 15 = (7x - 13) x 6

=> 30x +45 = 42x - 78

=> 30x - 42x = -78 - 45

=> -12x = -123

=> x = 41/4

d) 2-x/4 = 3x - 1/ - 3

=> (2-x) . -3 = 4. (3x - 1)

=> -6 - (-3x) = 12x - 4

(-6) + 4 = 12x - - (-3x)

=> -2 = 9x

=> x = -2/9    

1: x=3/4-1/2=3/4-2/4=1/4

2: x-1/5=2/11

=>x=2/11+1/5=21/55

3: x-5/6=16/42-8/56

=>x-5/6=8/21-4/28=5/21

=>x=5/21+5/6=15/14

4: x/5=5/6-19/30

=>x/5=25/30-19/30=6/30=1/5

=>x=1

5: =>|x|=1/3+1/4=7/12

=>x=7/12 hoặc x=-7/12

6: x=-1/2+3/4

=>x=3/4-1/2=1/4

11: x-(-6/12)=9/48

=>x+1/2=3/16

=>x=3/16-1/2=-5/16

1)x= 1/4

2)x= 2/11+ 1/5

   x= 21/55

3)x - 5/6 = 5/21

   x         = 5/21+5/6

   x         = 15/14

4)x/5 = 5/6 + -19/30

   x:5 = 1/5

   x    = 1/5.5

   x    = 1

5) |x| - 1/4 = 6/18

    |x|           = 6/18 - 1/4

    |x|            =7/12

⇒x= 7/12 hoặc -7/12

6)x = -1/2 +3/4

   x= 1/4

7) x/15 = 3/5 + -2/3

   x:15  = -1/15

  x        = -1/15. 15

  x        = -1

8)11/8 + 13/6 = 85/x  

       85/24      = 85/x

  ⇒      x           = 24

9) x - 7/8 = 13/12

   x          = 13/12 + 7/8

   x          = 47/24

10)x - -6/15 = 4/27  

     x            = 4/27 + (-6/15)

    x             = -34/135

11) -(-6/12)+x = 9/48

                    x= 9/48 - 6/12

                    x = -5/16

12) x - 4/6 = 5/25 + -7/15

      x -4/6  =  -4/15

     x           = -4/15 + 4/6

    x             = 2/5

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)

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

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

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)

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

\(\frac{2}{3}-x=-\frac{7}{6}\)

\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)

\(x=\frac{2}{3}+\frac{7}{6}\)

\(x=\frac{11}{6}\)

/ là phân số nha

11/13-(5/42-x)=(15/28-11/13)

11/13-(5/42-x)=-37/182

(5/42-x)=11/13+37/182

(5/42-x)=191/182

x=5/42-191/182

x=-254/273

vậy x=-254/273