Question

Tìm \(x\), biết :

a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5},\left(x\ne0\right)\)

b) \(x^{10}=25x^8\)

Answer 1

a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\\ x^8=x^7\\ \Rightarrow x=1;x=-1\)

b)\(x^{10}=25.x^8\\ x^2=25\\ \Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Answer 2

a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)

\(\Rightarrow x^8=x^7\)

\(\Rightarrow x^8-x^7=0\)

\(\Rightarrow x^7.x-x^7=0\)

\(\Rightarrow x^7\left(x-1\right)=0\)

\(\Rightarrow x-1=0\) (vì x^7 \(\ne\)0)

\(\Rightarrow\) x=1

b) x^10=25x^8

\(\Rightarrow x^8.x^2-25x^8=0\)

\(\Rightarrow x^8\left(x^2-25\right)=0\)

\(\Rightarrow x^8=0\) hoặc \(x^2-25=0\)

1) x^8=0

\(\Rightarrow\) x=0(1)

2) x^2 -25=0

x^2=0+25

x^2=25

x^2=5^2 hay x^2=(-5)^2

Suy ra x=5 hoặc x=-5 (2)

Từ (1) và (2)\(\Rightarrow\)x\(\in\left\{0;5;-5\right\}\)

EM KO CHÉP ĐÁP ÁN NHÉ