Question

Tìm x bt 

X^4=16

(X-2)^2x+3 =(x-2)^2x+1(x€z)

(X⁴+0,0001)(x+1995)=0

 

Answer 1

1.   x⁴=16

      x⁴ =2⁴

=)   x   =2

Answer 2

\(x^4=16\)

\(x^4=\left(\pm2\right)^4\)

\(x=\pm2\)

Answer 3

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

\(\left(x-2\right)^{2x+3}-\left(x-2\right)^{2x+1}=0\)

\(\left(x-2\right)^{2x+1}\left[\left(x-2\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{2x+1}=0\\\left(x-2\right)^2-1=0\end{cases}}\)                    \(\Rightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=0+1=1\end{cases}}\)                    \(\Rightarrow\orbr{\begin{cases}x=0+2=2\\\left(x-2\right)^2=\left(\pm1\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x-2=\pm1\end{cases}}\)                                   \(\Rightarrow\orbr{\begin{cases}x=2\\\begin{cases}x=-1+2=1\\x=1+2=3\end{cases}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1+2=1\text{ Hoặc }x=1+2=3\end{cases}}\)

                         \(\Rightarrow\text{ }x\in\left\{2\text{ ; }1\text{ ; }3\right\}\)

Answer 4

\(\left(x^4+0,0001\right)\left(x+1995\right)=0\)

\(\Rightarrow\text{ }\left(x+1995\right)=0\text{ ( Vì }x^4+0,0001\text{ }\ne0\text{ }\)

\(\Rightarrow\text{ }x=0-1995=-1995\)