Question

25 - y²=8 ( x - 2009 )

Tìm x,y

Answer 1

Đề có sai không vậy

Answer 2

\(25-y^2=8\left(x-2009\right)\)

\(\Leftrightarrow\)\(25-y^2=8x-16072\)

\(\Leftrightarrow\)\(8x=25-y^2-16072\)

\(\Leftrightarrow\)\(8x=25-16072-y^2\)

\(\Leftrightarrow\)\(8x=-16047-y^2\)

\(\Leftrightarrow\)\(x=\frac{-16047-y^2}{8}\)

Xét: \(8x=-16047-y^2\)

Thay vào: \(8\times\frac{-16047-y^2}{8}=-16047-y^2\)

\(\Leftrightarrow\)\(-160,47-y^2=-16047-y^2\)

\(\Leftrightarrow\)\(y\)có vô giá trị \(\left(y\in R\right)\)

Vậy \(\orbr{\begin{cases}x=\frac{-16047-y^2}{8}\\y\in R\end{cases}}\)

Answer 3

trả lời :

\(25-y^2=8\left(x-2009\right)\)

\(\Leftrightarrow\)\(25-y^2=8x-16072\)

\(\Leftrightarrow\)\(8x=25-y^2-16072\)

\(\Leftrightarrow\)\(8x=25-16072-y^2\)

\(\Leftrightarrow\)\(8x=-16047-y^2\)

\(\Leftrightarrow\)\(x=\frac{-16047-y^2}{8}\)

Xét: \(8x=-16047-y^2\)

Thay vào: \(8\times\frac{-16047-y^2}{8}=-16047-y^2\)

\(\Leftrightarrow\)\(-160,47-y^2=-16047-y^2\)

\(\Leftrightarrow\)\(y\)có vô giá trị \(\left(y\in R\right)\)

Vậy \(\orbr{\begin{cases}x=\frac{-16047-y^2}{8}\\y\in R\end{cases}}\)

^HT^