Question

Answer 1

Answer 2

b: Ta có: \(1-9x+27x^2-27x^3=0\)

\(\Leftrightarrow\left(1-3x\right)^3=0\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)

c: ta có: \(4x^2+9y^2=12xy\)

\(\Leftrightarrow\left(2x-3y\right)^2=0\)

\(\Leftrightarrow2x=3y\)

hay \(x=\dfrac{3}{2}y\)

Answer 3

c. 4x² + 9y² = 12xy

<=> (2x)² - 2.2x.3y + (3y)² = 0

<=> (2x + 3y)² = 0

<=> 2x + 3y = 0

<=> \(\left\{{}\begin{matrix}2x=-3y\\3y=-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3y}{2}\\y=\dfrac{-2x}{3}\end{matrix}\right.\)

Answer 4

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