7\(x\) - 2y = 15

y =( 7\(x\) - 15) : 2

⇒ 7\(x\) - 15 ⋮ 2

⇒ \(x\) - 1 ⋮ 2 

⇒ \(x\) = 2k + 1; k \(\in\) N

    Vì y là số tự nhiên nên 7\(x\) - 15 ≥ 0 ⇒ \(x\) ≥ \(\dfrac{15}{7}\) 

  ⇒ 2k + 1 ≥ \(\dfrac{15}{7}\)

       k ≥ (\(\dfrac{15}{7}\) - 1 ) : 2

     k ≥ \(\dfrac{8}{14}\) ⇒ k ≥ 1;

⇒ \(x\) = 2k + 1; k ϵ N*

y = \(\dfrac{7.\left(2k+1\right)-15}{2}\)

y = 7k - 4

Vậy câc cặp số tự nhiên \(x;y\) thỏa mãn đề bài là

(\(x;y\)) = (2k+1; 7k - 4); k \(\in\)N*

\(\left(\dfrac{-3}{4}\right)^2.\left(\dfrac{-6}{9}\right)^2\)

\(=\left(\dfrac{-3}{4}.\dfrac{-6}{9}\right)^2\)

\(=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

*Công thức: \(a^m.b^m=\left(a.b\right)^m\)

\(#Thitottt:3\)

\(CTTQ:H_xS_y\left(x,y:nguyên,dương\right)\\ M_{H_xS_y}=17M_{H_2}=17.2=34\left(\dfrac{g}{mol}\right)\\ x=\dfrac{5,88\%.34}{1}=2;y=\dfrac{34-2.1}{32}=1\\ \Rightarrow CTPT:H_2S\)

=1/4

=1/4 bn nhé

(-4)² x (-3/2)²

= 16 x (-1,5)²

= 16 x 2,25

= 36

(-4)².(-\(\dfrac{3}{2}\))²

= (- 4. (-\(\dfrac{3}{2}\)))²

= 6²

= 36

\(\left(-5\right)^3:\left(-5\right)^2=\left(-5\right)^{3-2}=-5\)

(-5)³ : (-5)² = (-5)^{3 - 2} = (-5)¹

\(2^3.2^4=2^{3+4}=2^7\)

2³ . 2⁴ = 2^{3 + 4} = 2⁷