Question

Tìm x: \(\left(2x-1\right)^8=\left(2x-1\right)^{10}\)

Answer 1

(2x - 1)⁸ = (2x - 1)¹⁰

(2x - 1)¹⁰ - (2x - 1)⁸ = 0

(2x - 1)⁸.[(2x - 1)² - 1] = 0

(2x - 1)⁸ = 0 hoặc (2x - 1)² - 1 = 0

*) (2x - 1)⁸ = 0

2x - 1 = 0

2x = 1

x = 1/2

*) (2x - 1)² - 1 = 0

(2x - 1)² = 1

2x - 1 = 1 hoặc 2x - 1 = -1

**) 2x - 1 = 1

2x = 2

x = 1

**) 2x - 1 = -1

2x = 0

x = 0

Vậy x = 0; x = 1/2; x = 1

Answer 2

(2x - 1)⁸ = (2x - 1)¹⁰

=) (2x - 1)¹⁰ : (2x - 1)⁸ = 1

    (2x - 1)² = 1 =) = 1²

=) 2x - 1 = 1

    2x = 2

      x = 1.

Answer 3

 các bước giải:

${(2x-1)}^8={(2x-1)}^{10}$
$\Rightarrow {(2x-1)}^8-{(2x-1)}^{10}=0$
$\Rightarrow {(2x-1)}^8.1-{(2x-1)}^8.{(2x-1)}^2=0$
$\Rightarrow {(2x-1)}^8.[1-{(2x-1)}^2]=0$
TH$1$
${(2x-1)}^8=0$
$\Rightarrow 2x-1=0$
$\Rightarrow 2x=0+1$
$\Rightarrow 2x=1$
$\Rightarrow x=1:2$
$\Rightarrow x=\frac{1}{2}$
TH$2$
$1-{(2x-1)}^2=0$
$\Rightarrow {(2x-1)}^2=1$
$\Rightarrow {(2x-1)}^2={(\pm 1)}^2$
$\Rightarrow 2x-1=1$ hoặc $2x-1=-1$
$\Rightarrow 2x=2$ hoặc $2x=0$
$\Rightarrow x=1$ hoặc $x=0$
Vậy $x\in \{\frac{1}{2};1;0\}$

Answer 4

cách bước giải

${(2x-1)}^8={(2x-1)}^{10}$
$\Rightarrow {(2x-1)}^8-{(2x-1)}^{10}=0$
$\Rightarrow {(2x-1)}^8.1-{(2x-1)}^8.{(2x-1)}^2=0$
$\Rightarrow {(2x-1)}^8.[1-{(2x-1)}^2]=0$
TH$1$
${(2x-1)}^8=0$
$\Rightarrow 2x-1=0$
$\Rightarrow 2x=0+1$
$\Rightarrow 2x=1$
$\Rightarrow x=1:2$
$\Rightarrow x=\frac{1}{2}$
TH$2$
$1-{(2x-1)}^2=0$
$\Rightarrow {(2x-1)}^2=1$
$\Rightarrow {(2x-1)}^2={(\pm 1)}^2$
$\Rightarrow 2x-1=1$ hoặc $2x-1=-1$
$\Rightarrow 2x=2$ hoặc $2x=0$
$\Rightarrow x=1$ hoặc $x=0$
Vậy $x\in \{\frac{1}{2};1;0\}$