a, 5 + x = 3

          x = 3 - 5

          x = 3 + (-5)

          x = -2

b, 15x = -75

       x = -75 : 15

       x = -5

c, 3x + 7 = x + 15

    3x - x  =  15 - 7

        2x   =    8 

          x   = 8 : 2

          x   = 4

d, lx-3l + 5 = 18

    lx-3l       = 18 - 5

    lx-3l       =  13

Suy ra x-3 =13 hoac x-3 = -13

TH1:

x - 3 = 13

x      = 13 + 3

x      = 16

TH2:

x - 3 = -13

x      = -13 + 3

x      = -10

k mk nha bn

A ) x = -2

B) x = -5

C) x =  4

D) x = 16

Nhớ tk cho mk nhé !

A. \(5+x=3\)

             \(x=3-5\)

              \(x=-2\)

B. \(15x=-75\)

          \(x=\left(-75\right)\div15\)

          \(x=-5\)

C. \(3x+7=x+15\)

           \(3x-x=15-7\)

                 \(2x=8\)

              \(x=8:2=4\)

D. \(\left|x-3\right|+5=18\)

\(\left|x-3\right|=18-5\)

\(\left|x-3\right|=13\)

\(\Rightarrow\orbr{\begin{cases}x-3=13\Rightarrow x=13+3=16\\x-3=-13\Rightarrow x=\left(-13\right)+3=-10\end{cases}}\)

-15x = -50

=> x = -50 : -15

=> x=10/3

-15x = -75 + (-5)²

-15x = -75 + 25

-15x = -50

    x = -50 : -15

    x = 10/3

k cho tớ nha

-15 x =-75 + (-5) ^2

-15 x =-75 + 25

-15 x =-50

    x  = (-50) : (-15)

    x  =3,4  ( mk làm tròn luôn nha)

Đặt x/2=y/5=k

=>x=2k; y=5k

xy-15x+6y=40

\(\Leftrightarrow10k^2-15\cdot2k+6\cdot5k=40\)

\(\Leftrightarrow10k^2=40\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

=>x=4;y=10

TRường hợp 2: k=-2

=>x=-4; y=-10

Đặt `x/2 = y/5 = k`

`=>` `{(x = 2k),(y = 5k):}`

Ta có `: xy - 15x + 6y = 40`

`=> 2k . 5k - 15 . ( 2k ) + 6 . ( 5k ) = 40`

`=> 10k^2 - 30k + 30k = 40`

`=> k^2 = 40 : 10`

`=> k^2 = 4`

`=>` \(\left[ \begin{array}{l}k^2 = 2^2\\k^2 = ( - 2 )^2\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}k = 2\\k = - 2\end{array} \right.\) 

Xét `k = 2 => {(x = 2 . 2 = 4),(y = 5 . 2 = 10):}`

Xét `k = - 2 => {(x = - 2 . 2 = - 4),(y = - 2 . 5= - 10):}`

Vậy `, ( x ; y ) in { ( 4 ; 10 ) ; ( - 4 ; - 10 ) } .`

` x/2=y/5 => x=2y /5`

thay` x = 2y /5 `vào `xy - 15x + 6y = 40` ta đc

\(\dfrac{2}{5}y^2-6y+6y=40\)

\(=>\dfrac{2}{5}y^2=40\)

\(=>y^2=100\Leftrightarrow\left[{}\begin{matrix}y=10\\y=-10\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{5}.10=4\\x=\dfrac{2}{5}.\left(-10\right)=-4\end{matrix}\right.\)

5^x + 5^x+1 = 75

=> 5^x ( 1 + 5 )

a) -15x + 20 =-25

\(\Leftrightarrow\) -15x = -25 - 20 = -45

\(\Leftrightarrow\)x = 3