\(\frac{x}{4}=\frac{25}{x}\)

\(\Rightarrow x^2=100\)

\(x^2=\left(\pm10\right)^2\)

\(x=\pm10\)

\(\frac{y2}{3}=\frac{12}{1}\)

\(y2=36\)

\(y=36:2\)

\(y=18\)

chúc bạn học tốt

Lời giải:

a. 

$(25-2x)^3:5-3^2=4^2$

$(25-2x)^3:5=4^2+3^2=25$

$(25-2x)^3=25.5=5^3$

$\Rightarrow 25-2x=5$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

b.

$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$

$\Rightarrow 3^x=3^{14}$

$\Rightarrow x=14$

a) x = 8.

b) x = -5.

c) x = 5.

d) x = -3.

− 19 + 4 ( 2 − x ) = 25 − x − 19 + 8 − 4 x = 25 − x   4 x − x = − 19 + 8 − 25   3 x = − 36 = > x = − 12  

/x+2/=0\(\Rightarrow x+2=0\Rightarrow x=0-2=-2\)

/x-3/=7-(-2)=9\(\Rightarrow x=9+3=12ho\text{ặc}x=-9+3=-6\)

(7-x)-(25+7)=-25\(\Rightarrow\left(7-x\right)-32=-25\Rightarrow7-x=-25+32=7\Rightarrow x=0\)

/x-3/=/5/+/-7/=5+7=12\(\Rightarrow x=15ho\text{ặc}x=-9\)

/x-5/=/-7/=7\(\Rightarrow x=12ho\text{ặc}x=-2\)

4-(7-x)=x-(13-4)\(\Rightarrow x-3=x-9\Rightarrow x-x=-9+3\Rightarrow0=-6\)(vô lí)

Vậy không có x thoả mãn 4-(7-x)=x-(13-4)

|x+2|=0

Vì |0|=0,suy ra x+2=0

                         x=0-2

                         x=-2

Vậy x = -2

|x-3|=7-(-2)

|x-3|=7+2

|x-3|=9

Vì |9|=|-9|=9,suy ra x-3 thuộc{9;-9}

*x-3=9

x   =9+3

x   =12

*x-3=-9

    x=-9+3

     x=-6

Vậy x thuộc {12;-6}

(7-x)-(25+7)=-25

(7-x)-32      =-25

            7-x =-25+32

            7-x  =57

              x   =7-57

              x   =7+(-57)

              x    =-50

vậy x=-50

|x-3|=|5|+|-7|

|x-3|=5+7

|x-3|=12

Vì |12|=|-12|=12,suy ra x-3 thuộc {12;-12}

*x-3=12

  x=12+3

  x=15

*x-3=-12

    x=-12+3

    x=-9

vậy x thuộc {15;-9}

x − 4 x 2 − 25 = 0 = > x − 4 = 0 x 2 − 25 = 0 = > x = 4 x 2 = 25 = > x = 4 x = ± 5