\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

a: \(\left(2x-3\right)\left(3x^2+1\right)-6x\left(x^2-x+1\right)+3x^2-2x=10\)

\(\Leftrightarrow6x^3+2x-9x^2-3-6x^3+6x^2-6x+3x^2-2x=10\)

\(\Leftrightarrow-6x-3=10\)

=>-6x=13

hay x=-13/6

b: \(\Leftrightarrow3x^2-3x+x-2-3x^2+5x=-8-5x\)

=>3x-2=-5x-8

=>8x=-6

hay x=-3/4

c: \(\Leftrightarrow64x^3-27-64x^3+32x^2-32x^2+x=20\)

=>x-27=20

hay x=47

a: Ta có: \(2^{x-1}=32\)

\(\Leftrightarrow x-1=5\)

hay x=6

b: Ta có: \(3^{2x+1}=81\)

\(\Leftrightarrow2x+1=4\)

\(\Leftrightarrow2x=3\)

hay \(x=\dfrac{3}{2}\)

c: Ta có: \(2^x-26=6\)

\(\Leftrightarrow2^x=32\)

hay x=5

d: Ta có: \(27\cdot3^x=243\)

\(\Leftrightarrow3^x=9\)

hay x=2

\(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Rightarrow\dfrac{2x+19}{21}-\dfrac{2x+17}{23}-\dfrac{2x+7}{33}+\dfrac{2x+5}{35}=0\)

\(\Rightarrow\left(\dfrac{2x+19}{21}+1\right)-\left(\dfrac{2x+17}{23}+1\right)-\left(\dfrac{2x+7}{33}+1\right)+\left(\dfrac{2x+5}{35}+1\right)=0\)

\(\Rightarrow\dfrac{2x+40}{21}-\dfrac{2x+40}{23}-\dfrac{2x+40}{33}+\dfrac{2x+40}{35}=0\)

\(\Rightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Rightarrow2x+40=0\Rightarrow x=-20\)( do \(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}>0\))

Ta có: \(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Leftrightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow2x+40=0\)

hay x=-20

a. 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$

$\Leftrightarrow \sqrt{2x}=3$

$\Leftrightarrow 2x=9$

$\Leftrightarrow x=\frac{9}{2}$ (tm)

b.

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$

$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$

$\Leftrightarrow 3\sqrt{x+2}=15$

$\Leftrightarrow \sqrt{x+2}=5$

$\Leftrightarrow x+2=25$

$\Leftrightarrow x=23$ (tm)

c.

$\sqrt{(x-2)^2}=12$

$\Leftrightarrow |x-2|=12$

$\Leftrightarrow x-2=12$ hoặc $x-2=-12$

$\Leftrightarrow x=14$ hoặc $x=-10$

e.

PT $\Leftrightarrow |2x-1|-x=3$

Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

f.

ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$

$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$

$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$

$\Leftrightarrow x=2$ hoặc $x=5$ (tm)

h. ĐKXĐ: $x\leq \frac{3}{2}$

PT $\Leftrightarrow \sqrt{3-2x}=x+2$

\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)

\(\Leftrightarrow x=-3+2\sqrt{2}\) (tm)

Vậy.......

a: \(x^2\left(2x-3\right)+8x-12=0\)

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

=>2x-3=0

hay x=3/2

b: \(\Leftrightarrow\left(2x-5\right)\left(2x+10\right)-\left(2x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+10-x+1\right)=0\)

=>(2x-5)(x+11)=0

=>x=5/2 hoặc x=-11

c: \(\Leftrightarrow2x\left(x^2-16\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)

hay \(x\in\left\{0;4;-4\right\}\)

a) 2x = 16 <=>x=8

b) 3x+1 = 9x <=>9x-3x=1

<=>6x=1 <=>x=1/6

c) 23x+2 = 4x+5 <=>23x-4x=5-2

<=>19x=3 <=>x=3/19

d) 32x-1 = 243 <=>32x=244

<=>x=61/8

a/ 2x=16

x=8

b/ 3x+1=9x

3x-9x=-1

-6x=-1

x=1/6

c/ 23x+2=4x

23x-4x=-2

19x=-2

x=-2/19

d/ 32x-1=243

32x=244

x=61/8