a, `(x-9)^4=(x-9)^7`
`(x-9)^4-(x-9)^7=0`
`(x-9)^4 . [(1-(x-9)^3]=0`
TH1: `(x-9)^4=0`
`x-9=0`
`x=9`
TH2: `1-(x-9)^3=0`
`(x-9)^3=1^3`
`x-9=1`
`x=10`
b, `(3x-15)^10=(3x-15)^15`
`(3x-15)^10 . [1-(3x-15)^5]=0`
TH1: `(3x-15)^10=0`
`3x-15=0`
`x=5`
TH2: `1-(3x-15)^5=0`
`(3x-15)^5=1^5`
`3x-15=1`
`x=16/3` (Loại)
c, `(x-8)^3=(x-8)^6`
`(x-8)^3 .[1-(x-8)^3]=0`
TH1: `(x-8)^3=0`
`x=8`
TH2: `1-(x-8)^3=0`
`x-8=1`
`x=9`
\(a,\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left(x-9\right)=\left(x-9\right)^2\)
\(\Rightarrow\left(x-9\right)^3\)
\(\Rightarrow x=9\)
\(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left(3x-15\right)=\left(3x-15\right)^5\)
\(\Rightarrow\left(3x-15\right)^6\)
\(\Rightarrow3x-15=0\)
\(3x=15\)
\(x=15:3\)
\(x=...\)
Mấy phần kia bn có thể áp dụng gần giống ntn !
\(c,\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left(x-8\right)^9\)
\(\Rightarrow x-8=0\)
\(\Rightarrow x=8\)
a: Ta có: \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Leftrightarrow\left(x-9\right)^4\left[\left(x-9\right)^3-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
b: Ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)
\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)
c: Ta có: \(\left(x-8\right)^6=\left(x-8\right)^3\)
\(\Leftrightarrow\left(x-8\right)^3\cdot\left[\left(x-8\right)^3-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
