a) \(\left(\left|x\right|+3\right):5-3=12\Leftrightarrow\left|x\right|+3=45\Leftrightarrow\orbr{\begin{cases}x=42\\x=-42\end{cases}}\)

b) \(86:\left[2\left(2x-1\right)^2-7\right]+4^2=2\cdot3^2\Leftrightarrow2\left(2x-1\right)^2-7=43\Leftrightarrow\left(2x-1\right)^2=25\Leftrightarrow\orbr{\begin{cases}2x-1=-5\\2x-1=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

a) \(\left(\left|x\right|+3\right)\div5-3=12\)

    \(\left(\left|x\right|+3\right)\div5=12+3\)

    \(\left(\left|x\right|+3\right)\div5=15\)

       \(\left|x\right|+3=15.5\)

       \(\left|x\right|+3=75\)

       \(\left|x\right|=75-3\)

       \(\left|x\right|=72\)

\(\Rightarrow\orbr{\begin{cases}x=72\\x=-72\end{cases}}\)

Vậy \(x\in\left\{72;-72\right\}\)

b) \(86\div\left[2,\left(2x-1\right)^2-7\right]+4^2=2.3^2\)

    \(86\div\left[2.\left(2x-1\right)^2-7\right]+16=18\)

    \(86\div\left[2.\left(2x-1\right)^2-7\right]=18-16\)

    \(86\div\left[2.\left(2x-1\right)^2-7\right]=2\)

                  \(2.\left(2x-1\right)^2-7=86\div2\)

                  \(2.\left(2x-1\right)^2-7=43\)

                  \(2.\left(2x-1\right)^2=43+7\)

                  \(2.\left(2x-1\right)^2=50\)

                      \(\left(2x-1\right)^2=50\div2\)

                      \(\left(2x-1\right)^2=25\)    

                      \(\left(2x-1\right)^2=5^2\)

\(\Rightarrow2x-1=5\)

                \(2x=5+1\)

                \(2x=6\)

                  \(x=6\div2\)

                  \(x=3\)

a, (|x| +3):5-3=12

(|x|+3):5=15

|x|+3=3

|x|=0

=>x=0

a,10x+65=125

   10x      =125-65

       x      =60:10

       x      =6

Vậy:...

b,45-(5-2x)^3=2.3^2

   45-(5-2x)^3=18

    (5-2x)^3    =45-18

    (5-2x)^3    =27=3^3

=> 5-2x  = 3

     2x = 5-3

     x= 2:2

     x=1

Vậy:....

-Hok tốt ~ K mk nhak.

a, 10x + 65 = 125

10x = 125 - 65

10x = 60

x = 60 : 10

x = 6

b,45-(5-2x)^3=2.3^2

45 - ( 5 - 2x )^3 =18

(5 - 2x)^3 = 45 - 18

(5 - 2x )^3= 3^3

5 - 2x = 3

2x = 2

x = 1

c,2(x-3)-12=(-10)

2x - 6 -12 =-10

2x = -10 + 6 + 12

2x = 8

x = 4

d,x-12=(-13)+1+|-13|

x - 12 = -13 + 1 + 13

x - 12 = 1

x = 13

hok tốt!!!

\(\text{a)}\)\(10x+65=125\)

\(\Rightarrow10x=60\)

\(\Rightarrow x=6\)

\(b\text{)}\)\(45-\left(5-2x\right)^3=2.3^2\)

\(\Rightarrow45-\left(5-2x\right)^3=18\)

\(\Rightarrow\left(5-2x\right)^3=45-18\)

\(\Rightarrow\left(5-2x\right)^3=27\)

\(\Rightarrow\left(5-2x\right)^3=3^3\)

\(\Rightarrow5-2x=3\)

\(\Rightarrow2x=2\Rightarrow x=1\)

 giúp mình nhanh với 

1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)

2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)

\(2^x=2.2^8=2^9;x=9\)

4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)

\(\Leftrightarrow3^5.x=3^5.5;x=5\)

1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

$\Leftrightarrow x^3(x^2-1)=0$

$\Leftrightarrow x^3(x-1)(x+1)=0$

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

$(2x-1)^2-(2x-1)=0$

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

\(x^5-x^3=0\)

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

\((2x-1)^2-(2x-1)=0\)

`=> (2x-1)(2x-1) - (2x-1)=0`

`=> (2x-1)(2x-1-1)=0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

a) \(2^{2x}.2^4=1024\)

\(2^{2x}=1024:2^4\)

\(2^{2x}=1024:16\)

\(2^{2x}=64\)

\(2^{2x}=2^6\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

vay \(x=3\)

b) \(2.3^x=10.3^{12}+8.27^4\)

\(2.3^x=2.5.3^{12}+2^3.\left(3^3\right)^4\)

\(2.3^x=2.5.3^{12}+2^3.3^{12}\)

\(2.3^x=2.3^{12}.\left(5+2^2\right)\)

\(2.3^x=2.3^{12}.9\)

\(2.3^x=2.3^{12}.3^2\)

\(2.3^x=2.3^{14}\)

\(\Rightarrow x=14\)

vay \(x=14\)

c) \(5^8.25^x+1=5^{17}\)

\(5^8.\left(5^2\right)^x+1=5^{17}\)

\(5^8.5^{2x}+1=5^{17}\)

\(5^{8+2x}=5^{17}-1\)

e) \(\left(2x-4\right)^5=\left(2x-4\right)^3\)

\(\left(2x-4\right)^5-\left(2x-4\right)^3=0\)

\(\left(2x-4\right)\left[\left(2x-4\right)^2-1\right]=0\)

\(\left(2x-4\right)\left(2x-4-1\right)\left(2x-4+1\right)=0\)

\(\left(2x-4\right)\left(2x-5\right)\left(2x-3\right)=0\)

\(\Rightarrow2x-4=0\)hoac  \(\orbr{\begin{cases}2x-5=0\\2x-3=0\end{cases}}\)

\(\Rightarrow2x=4\)hoac \(\orbr{\begin{cases}2x=5\\2x=3\end{cases}}\)

 \(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)

              vay \(x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)