1: Tìm x

a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)

\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)

\(\Leftrightarrow3^x=3^{12}\cdot3^3\)

\(\Leftrightarrow3^x=3^{15}\)

hay x=15

Vậy: x=15

b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)

\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)

Vậy: x=2

a: =>x/27+1=-2/3

=>x/27=-5/3

=>x=-45

b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)

=>x=221/50

c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)

=>x=1/15-2/3=1/15-10/15=-9/15=-3/5

d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)

=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)

=>x=-37/24

e: =>-3/7x=84/45

=>x=-196/45

f: =>11/10x=-2/3

=>x=-20/33

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

Bài 2:

a: =>2/3x=3/4+1/2=3/4+2/4=5/4

=>x=5/4:2/3=5/4*3/2=15/8

b:=>-2x+4=3x-12

=>-5x=-16

=>x=16/5

a,3/4 . (-5/12)+3/4.(-7/12)

` 3/4 . [ - ( 5/12 + 7/12  ) ] `

`3/4 . (-1) = -3/4 `

`2/3 . x - 0,5 = 3/4 `

`        x - 0,5 = 3/4 - 2/3 `

`        x-0,5 = 1/12 `

`        x =       1/12 + 0,5 `

`        x= 7/12 `

a, 3/4 . (-5/12)+3/4.(-7/12) 

= 3/4 . [(-5/12) + (-7/12)]

= 3/4 . (-1)

= -3/4

----------------------------------------------------------------------------

a, 2/3 .x-0,5=3/4

2/3 . x = 3/2 + 0,5

2/3 . x = 2 

x = 2 : 2/3

x = 3 

vậy x = 3

a) (125 - x) + 25 = 98

125 - x = 98 - 25

125 - x = 73

x = 125 - 73

x = 52

b) 100 : (x + 20) = 68 : 34

100 : (x + 20) = 2

x + 20 = 100 : 2

x + 20 = 50

x = 50 - 20

x = 30

c) (3x + 12) . 5 = 25 . 4 - 10

(3x + 12) . 5 = 100 - 10

(3x + 12) . 5 = 90

3x + 12 = 90 : 5

3x + 12 = 18

3x = 18 - 12

3x = 6

x = 6 : 2

x = 3

d) 210 : (2x - 3) - 20 = 10

210 : (2x - 3) = 10 + 20

210 : (2x - 3) = 30

2x - 3 = 210 : 30

2x - 3 = 70

2x = 70 + 3

2x = 73

x = 73/2

a, ( 125 - x ) + 25 = 98

⇒ 125 - x = 73

⇒ x = 52.

Vậy..

b, 100 : ( x + 20 ) = 68 : 34

⇒ 100 : ( x + 20 ) = 2

⇒ x + 20 = 50

⇒ x = 30

Vậy...

c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10

⇒ ( 3 . x + 12 ) . 5 = 90

⇒ 3 .x + 12 = 18

⇒ 3x = 6

⇒ x = 2.

Vậy...

d, 210 : ( 2 .x - 3 ) - 20 = 10

⇒ ( 2 .x - 3 ) - 20 = 21

⇒ 2x - 3 = 41

⇒ 2x = 44

⇒ x = 22.

Vậy..

Cho mk chỉnh lại câu cuối:

 d, 210 : ( 2 .x - 3 ) - 20 = 10

⇒ 210 : ( 2 .x - 3 ) = 30

⇒ 2 . x - 3 = 7

⇒ 2x = 10

⇒ x = 5.

Vậy..

11: Ta có: \(\left(x+3\right)^3=125\)

\(\Leftrightarrow x+3=5\)

hay x=2

12: Ta có: \(\left(2x\right)^4=16\)

\(\Leftrightarrow x^4=1\)

hay \(x\in\left\{1;-1\right\}\)

11: Ta có: \left(x+3\right)^3=125(x+3)3=125

$\iff x+3=5$

hay x=2

12: Ta có: \left(2x\right)^4=16(2x)4=16

\Leftrightarrow x^4=1(2x)   = 2   =2  4

           4      

⇔2x=2⇔ x=1

13) 32:(3x-2)=2^3

32:(3x-2)=8

3x-2=32:8=4

3x=4+2=6

x=6:3=2

⇔x=2