\((2x-3)^3=(2x-3)^5\\\Rightarrow (2x-3)^3-(2x-3)^5=0\\\Rightarrow (2x-3)^3[1-(2x-3)^2]=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x-3=1\\2x-3=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\2x=4\\2x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

\(Vậy:x\in\left\{1;\dfrac{3}{2};2\right\}\)

`#3107.101107`

`|3x - 1| = x + 2`

`\Rightarrow` TH1: `3x - 1 = x + 2`

`\Rightarrow 3x - x = 2 + 1`

`\Rightarrow 2x = 3`

`\Rightarrow x =` $\dfrac{3}2$

TH2: `3x - 1 = -(x + 2)`

`\Rightarrow 3x - 1 = -x - 2`

`\Rightarrow 3x + x = -2 + 1`

`\Rightarrow 4x = -1`

`\Rightarrow x =` $\dfrac{-1}4$

Vậy, \(x\in\left\{-\dfrac{1}{4};\dfrac{3}{2}\right\}.\)

|3x - 1| = x + 2

*) TH1: x ≥ 1/3, ta có:

|3x - 1| = x + 2

3x + 1 = x + 2

3x - x = 2 - 1

2x = 1

x = 1/2 (nhận)

*) TH2: x < 1/3, ta có:

|3x - 1| = x + 1

1 - 3x = x + 1

-3x - x = 1 - 1

-4x = 0

x = 0 (nhận)

Vậy x = 0; x = 1/2

`#3107.101107`

\(\left(-27\right)^5\div32^3\\ =\left[\left(-3\right)^3\right]^5\div\left(2^5\right)^3\\ =\left(-3\right)^{15}\div2^{15}\\ =\left(-3\div2\right)^{15}\\ =\left(-\dfrac{3}{2}\right)^{15}\)

chịu luôn!

Lời giải:
$M$ là trung điểm của $CD$ nên $CM=\frac{1}{2}CD$

$\Rightarrow 15=\frac{1}{2}CD$

$\Rightarrow CD=30$ (cm)

Vì M là trung điểm của CD nên CM = MD = 15 (cm)

Do đó CD = CM + MD = 15 + 15 = 30 (cm)