Question

9(x - 1 )^2 - (x - 1)^4 = 0

 (2x - 3 )^2 + 2x = 3

C.onn mng nhìuu ạ

Answer 1

\(9\left(x-1\right)^2-\left(x-1\right)^4=0\\ \Rightarrow\left(x-1\right)^2\left[9-\left(x-1\right)^2\right]\\ \Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\9-\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\9=\left(x-1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\3^2=x-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x-1=3\\x-1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-2\end{matrix}\right.\) 

________________________

\(\left(2x-3\right)^2+2x=3\\ \Rightarrow\left(2x-3\right)^2+\left(2x-3\right)=0\\ \Rightarrow\left(2x-3\right)\left(2x-3+1\right)=0\\ \Rightarrow\left(2x-3\right)\left(2x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\2x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\2x=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{2}{2}=1\end{matrix}\right.\)

Answer 2

đừng đăng linh tinh nhé bạn (Phạm Minh Khuê)
 

Answer 3

9.(\(x-1\))² - (\(x-1\))⁴ = 0

   (\(x-1\))²(9 - (\(x-1\))²) = 0

    \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\9-\left(x-1\right)^2=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=1\\x-1=-3\\x-1=3\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=1\\x=-3+1\\x=3+1\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=1\\x=-2\\x=4\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- 2; 1; 4}

Answer 4

(2\(x\) - 3)² + 2\(x\) = 3

(2\(x\) - 3)² + 2\(x\) - 3 = 0

(2\(x-3\))(2\(x\) - 3 + 1) = 0

(2\(x-3\))(2\(x-2\)) = 0

\(\left[{}\begin{matrix}2x-3=0\\2x-2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=3\\2x=2\end{matrix}\right.\)

 \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)

Vậy \(x\in\) {1; \(\dfrac{3}{2}\)}