a) (2x-3)¹⁵ = (2x-3)⁷

=> (2x-3)¹⁵ - (2x-3)⁷ = 0

(2x-3)⁷.[(2x-3)⁸ -1] = 0

=> (2x-3)⁷ = 0 => 2x-3 = 0 => 2x = 3 => x = 3/2

(2x-3)⁸ - 1 = 0 => (2x-3)⁸ = 1 => 2x - 3 = 1 => 2x = 4 => x = 2

                                              => 2x - 3 = - 1 => 2x = 2 => x = 1

KL:...

b) ta có: \(\left(x-3\right)^{16}\ge0;\left(3y-5\right)^4\ge0.\)

Để (x-3)¹⁶ + (3y-5)⁴ = 0

=> (x-3)¹⁶ = 0 => x-3 = 0 => x = 3

(3y-5)⁴ = 0 => 3y - 5 = 0 => 3y = 5 => y = 5/3

KL:...

= 1111

1000+1+110= 1111

Rồi để ảnh tớ đi bạn!~

\(\frac{x-3}{x-5}\)\(=\frac{5}{7}\)

\(\Rightarrow\frac{x}{x}\)\(=\frac{5-3}{7-5}\)

\(\Rightarrow\frac{x}{x}\) \(=\frac{2}{2}\)

Vậy \(x=2\)

\(\frac{x-3}{x-5}=\frac{5}{7}\)

<=> (x - 3) . 7 = (x - 5) . 5

<=> 7x - 21 = 5x - 25

<=> 7x = 5x - 25 + 21

<=> 7x = 5x - 4

<=> 7x = 5x - 4 - 5x

<=> 2x = -4

<=> x = -4 : 2

<=> x = -2

=> x = -2

\(\left(x+2\right)^3=125=5^3\)

\(\Rightarrow x+2=5\)

\(\Rightarrow x=3\)

Vậy,...........

OK nice easy 

\(\left(x+2\right)^3=125=5^3\)

\(\Rightarrow x+2=5\)

\(\Rightarrow x=3\)

\(5\cdot2^x+4\cdot2^x=72\)

\(2^x\cdot\left(5+4\right)=72\)

\(2^x=72\div9\)

\(2^x=8=2^3\)

\(\Rightarrow x=3\)

5.\(2^x\)+4.\(2^x\)=72

(=)\(2^x\).(5+4)=72

(=)\(2^x\)=8

(=)\(2^x=2^3\)

=>x=3

\(3^{x+2}-3^x=72\)

\(3^x\cdot3^2-3^x=72\)

\(3^x\cdot\left(3^2-1\right)=72\)

\(3^x=9=3^2\)

\(\Rightarrow x=2\)

3^{x + 2} - 3^x = 72

3^x(3² - 1) = 72

3^x , 8 = 72

3^x = 9

x = 2

Festivals around the world : Lễ hội trên toàn thế giới .

Mk ko vt 1 đoạn văn để giúp bn đk mong bn thông cảm nhé .

# MissyGirl #

\(xy=\frac{x}{y}\Leftrightarrow y=\frac{1}{y}\Rightarrow\orbr{\begin{cases}y=1\\y=-1\end{cases}}\)
Thay y vào pt x-y=xy để tìm x