Sửa: \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\le0\)

Mà \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\ge0\) với mọi x,y

Do đó \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{18}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=6+18=24\)

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=k\Leftrightarrow a=2k;b=3k\)

\(ab=24\Leftrightarrow6k^2=24\Leftrightarrow k^2=2\\ \Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4;b=6\\a=-4;b=-6\end{matrix}\right.\)

Ta có :

\(\dfrac{a}{2}=\dfrac{b}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\end{matrix}\right.\)

mà \(ab=24\)

\(\Rightarrow2k.3k=24\)

\(\Rightarrow6k^2=24\)

\(\Rightarrow k^2=2^2\)

\(\Rightarrow k=\left\{{}\begin{matrix}2\\-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}=2\\\dfrac{a}{2}=\dfrac{b}{3}=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=4;b=6\\a=-4;b=-6\end{matrix}\right.\)

Ta có:

(a.b)/(2.3)= 24/6=4(T/c dãy tỉ số = nhau )

=>a=4.2=8

b=4.3=12

6 x 8 = 48

5 x 6 = 30 

8 x 10 = 80

12 x 2 = 24

10 x 15 = 150

48

30

80

24

150

pn nha nếu đung tk cho mk nha

6x8=48

5x6=30

8x10=80

12x2=24

10x15=150

a) \(\dfrac{x}{3}=\dfrac{-10}{6}\)

\(x\times6=-10\times3\)

\(x\times6=-30\)

\(x=-5\)

b) \(\dfrac{-8}{x}=\dfrac{-9}{15}\)

\(x\times-9=15\times-8\)

\(x\times-9=-120\)

\(x=\dfrac{40}{3}\)

c) \(\dfrac{2,7}{0,9}=\dfrac{-8}{x}\)

\(x\times2,7=-8\times0,9\)

\(x\times2,7=-7,2\)

\(x=-\dfrac{8}{3}\)

d) \(\dfrac{4}{9}=\dfrac{x}{12}\)

\(x\times9=12\times4\)

\(x\times9=48\)

\(x=\dfrac{48}{9}\)

\(x=\dfrac{16}{3}\)

Tham khảo:

Tham khảo!

https://olm.vn/hoi-dap/detail/10185285054.html

Lời giải:
$3xy-2x-2y=24$

$\Rightarrow (3xy-2x)-2y=24$

$\Rightarrow x(3y-2)-2y=24$

$\Rightarrow 3x(3y-2)-6y=72$

$\Rightarrow 3x(3y-2)-2(3y-2)=76$

$\Rightarrow (3x-2)(3y-2)=76$

Vì $x,y$ nguyên nên $3x-2, 3y-2$ cũng là số nguyên. Do đo $3x-2, 3y-2$ là ước của 76. 

Đến đây thì đơn giản rồi. Bạn chỉ cần xét các TH khác nhau của ước của 76.

\(\Leftrightarrow12x^3-12x^3-8x=24\)

hay x=-3