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(x-2)6 = (x-2)10
<=> (x-2)6 - (x-2)10 = 0
<=> (x-2)6[1-(x-2)4] = 0
<=> \(\left[\begin{array}{nghiempt}\left(x-2\right)^6=0\\1-\left(x-2\right)^4=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x-2=0\\\left(x-2\right)^4=1\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=2\\\left[\begin{array}{nghiempt}x-2=1\\x-2=-1\end{array}\right.\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=2\\\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\end{array}\right.\)
Vậy x \(\in\){2;3;1}
1) <=> x+2=+-4 <=> x=-2 +-6
2) \(\left(x-2\right)^8=\left(x-2\right)^6\Leftrightarrow\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\Leftrightarrow\left(x-2\right)^6\left(x-2-1\right)\left(x-2+1\right)=0\Leftrightarrow\left(x-2\right)^6\left(x-3\right)\left(x-1\right)=0\)=> x=6 hoặc x=3 hoặc x=1
a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)
\(\Rightarrow x^8=x^7\)
\(\Rightarrow x^8:x^7=1\)
\(\Rightarrow x=1\)
Vậy x = 1
b) \(x^{10}=25.x^8\)
\(\Rightarrow x^{10}:x^8=25\)
\(\Rightarrow x^2=25\)
\(\Rightarrow x=\pm5\)
Vậy \(x=\pm5\)
x=2
vi \(6^2+8^2=10^2\)
tại sao bn lai bít x=2
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