Có: \(x^2+5\left(2y-2020\right)^2=100=10^2+5\cdot0^2\)

\(\Rightarrow\hept{\begin{cases}x^2=10^2\\\left(2y-2020\right)^2=0^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\pm10\\y=1010\end{cases}}\)

Vậy \(\left(x;y\right)\in\left\{\left(10;1010\right);\left(-10;1010\right)\right\}\)

Answer:

Mình sửa lại đề nhé: \(100-y^2 = 5(2x-2020)^2\)

Có: 

\(y^2\ge0\)

\(\Rightarrow100-y^2\le100\)

\(\Rightarrow5\left(2x-2020\right)^2\le100\)

\(\Rightarrow\left(2x-2020\right)^2\le20\)

Mà: \(\left(2x-2020\right)^2\) là số chính phương chẵn

\(\Rightarrow\left(2x-2020\right)^2\in\left\{0;4;16\right\}\)

\(\Rightarrow y^2\in\left\{100;80;20\right\}\) 

Mà: \(y^2\) là số chính phương

\(\Rightarrow\hept{\begin{cases}y^2=100\\\left(2x-2020\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}y=\pm10\\x=1010\end{cases}}\)

\(\frac{2-x}{4}=\frac{3x-1}{3}\)

\(\frac{3\left(2-x\right)}{12}=\frac{4\left(3x-1\right)}{12}\)

\(=6x-3=12x-4\)

\(=1=6x\)

\(x=\frac{1}{6}\)

tìm x o ai dùng dấu = ;dấu tương đương hoặc suy ra

\(\frac{2-x}{4}=\frac{3x-1}{3}\)

<=>\(\frac{3\left(2-x\right)}{12}=\frac{4\left(3x-1\right)}{12}\)

<=>  6-3x=12x-4

<=> -3x-12x=-6-4

<=>-15x=-10

<=>x=-10/-15

<=>x=5/3

Vậy x=5/3

Ta có  \(y=f\left(x\right)=3x^2+1\)

\(f\left(\frac{1}{2}\right)=3.\left(\frac{1}{2}\right)^2+1=3.\frac{1}{4}\)\(+1=\frac{7}{4}\)

\(f\left(1\right)=3.1^2+1=3+1=4\)

\(f\left(3\right)=3.3^2+1=3.9+1=27+1=28\)

dễ mà no bít lm:))

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