Question

[Nguyễn Bình Minh]

tim x thuoc Z biet :
(x-1)^2 =(x-3)^4
HELP ME:0!!

Answer 1

[HT.Phong (9A5)]

\(\left(x-1\right)^2=\left(x-3\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-3\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left[\left(x-3\right)^2\right]^2=0\)

\(\Leftrightarrow\left[\left(x-1\right)-\left(x-3\right)^2\right]\left[\left(x-1\right)+\left(x-3\right)^2\right]=0\)

\(\Leftrightarrow\left(x-1-x^2+6x-9\right)\left(x-1+x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(-x^2+7x-10\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy: ... 

Answer 2

[Đỗ Gia Phát]

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