Question

a, 4*(3-x)^10=9(3-x)^8

b, (x-1)^x+3=(x-1)^x+1

 

Answer 1

`4.(3-x)^10=9(3-x)^8`

`=>(3-x)^{8}(4(3-x)^2-9)=0`

`=>` $\left[ \begin{array}{l}3-x=0\\(3-x)^2=\dfrac{9}{4}\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\3-x=\dfrac{3}{2}\\3-x=-\dfrac{3}{2}\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=\dfrac{3}{2}\\x=\dfrac{9}{2}\end{array} \right.$

Vậy `x=3` hoặc `x=3/2` hoặc `x=9/2`

`b,(x-1)^{x+3}=(x-1)^{x+1}`

`=>(x-1)^{x+1}[(x-1)^2-1]=0`

`=>` $\left[ \begin{array}{l}x-=1\\x-=-1\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=2\\x=1\\x=0\end{array} \right.$

Vậy `x=0` hoặc `x=1` hoặc `x=2`

Answer 2

a) 4(3 - x)¹⁰ = 9(3 - x)⁸

=> (x - 3)⁸.[4(x - 3)² - 9) = 0

<=> (x - 3)⁸.(2x - 6 - 3)(2x - 6 + 3) = 0

<=> (x - 3)⁸(2x - 9)(2x - 3) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\2x-9=0\\2x-3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{9}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy S = {3;9/2; 3/2}

b) (x - 1)^{x + 3} = (x - 1)^{x  + 1}

<=> (x - 1)^{x + 1}.[(x - 3)² - 1) = 0

<=>(x - 1)^{x + 1}.(x - 3 - 1)(x - 3 + 1) = 0

<=> (x - 1)^{x + 1}.(x - 4)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=4\\x=2\end{matrix}\right.\)

Answer 3

b) Ta có: \(\left(x-1\right)^{x+3}=\left(x-1\right)^{x+1}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+1}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy: S={0;1;2}