123456578 (ahihi mik đùa hoi chứ mik hong có bít làm. Nhân tiện cảm ơn bạn vì lời động viên của bạn :D, mik cũng xin lỗi vì ko làm đc nhưng mik đã cố gắng hết sức rồi :'() 

!ERROR 404!

\(a^n=1\Rightarrow a^n=a^0\Rightarrow\left\{{}\begin{matrix}n=0\\a\in N\end{matrix}\right.\)

\(\left(3x-\dfrac{1}{3}\right)^2=16=4^2\)

\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{3}=4\\3x-\dfrac{1}{3}=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=4+\dfrac{1}{3}\\3x=-4+\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{13}{3}\\3x=-\dfrac{11}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\\x=-\dfrac{11}{9}\end{matrix}\right.\)

! LỖI MẠNG !

Bạn xem lại đề

B = \(\dfrac{x^2-2x+1}{x+1}\)

Với \(x\in\)Z, để B là số nguyên thì \(x^2-2x+1\)⋮ \(x+1\)

Theo Bezout ta có:   F(\(x\)) =  \(x^2\) - 2\(x\) + 1  ⋮ \(x+1\) ⇔ F(\(-1\))  ⋮ \(x+1\)

⇒ (-1)² - 2.(-1) + 1 ⋮ \(x\) + 1 ⇔ 4 ⋮ \(x\) + 1

⇔ \(x\) + 1 \(\in\) Ư(4) = { -4; -2; -1; 1; 2; 4}

\(\Leftrightarrow\) \(x\) \(\in\) { -5; -3; -2; 0; 1; 3}

Lời giải:
$7^{3x-2}-3.7^3=7^3.4$

$7^{3x-2}=3.7^3+7^3.4=7^3(3+4)=7^3.7=7^4$
$\Rightarrow 3x-2=4$

$\Rightarrow 3x=6$

$\Rightarrow x=2$

\(\left(3x-\dfrac{1}{3}\right)^2=16=4^2=\left(-4\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{3}=4\\3x-\dfrac{1}{3}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{13}{3}\\3x=-\dfrac{11}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\\x=-\dfrac{11}{9}\end{matrix}\right.\)

`(2x-1/3)^2=4`

\(< =>\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}:2\\x=-\dfrac{5}{3}:2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}\cdot\dfrac{1}{2}\\x=-\dfrac{5}{3}\cdot\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)