Question

Tìm x biết:

\(2^x = 4^{y-1} \) và \(27^y = 3^{x+8}\)

\((x^2 - 1)^2 + 1 = x^2\)

\((x - 1)^4 + 5.(y^2 - 1)^2 = 9\)

Answer 1

\(•\left(x^2-1\right)^2+1=x^2\\ \left(x^2-1\right)^2-x^2+1=0\\ x^4-2x^2+1-x^2+1=0\\ x^4-x^2-2x^2+2=0\\ \left(x^2-1\right)\left(x^2-2\right)=0\\ \left(x+1\right)\left(x-1\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x+\sqrt{2}=0\\x-\sqrt{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\)

Answer 2

Tìm y nữa