1) \(\Rightarrow x^2\left(x^{2004}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{2004}=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

2) \(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=\dfrac{86}{2}\\ x=43\)

\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=15^{10}:3^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=5^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=86:2\\ x=43\)

\(x^{20}=x\)

⇒\(x^{20}-x=0\)

⇒\(x\left(x^{19}-1\right)=0\)

⇔\(x^{19}-1=0\) hoặc x=0

⇔x=1 hoặc x=0

\(x^{20}=x\\ \Leftrightarrow x^{20}-x=0\\ \Leftrightarrow x\left(x^{19}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(x^{20}=x\\ \Rightarrow x\in\left\{0;1\right\}\)vì \(x\in N\)

gsyfdyfbfsyf

\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)

\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)

\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)

hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)

`#040911`

`(x + 5)^3 = (2x)^3`

`\Rightarrow x + 5 = 2x`

`\Rightarrow x + 5 - 2x = 0`

`\Rightarrow 5 + (x - 2x) = 0`

`\Rightarrow 5 - x = 0`

`\Rightarrow x = 5 - 0`

`\Rightarrow x = 5`

Vậy, `x= 5.`

Cảm ơn bạn nha