Question

[Cầm Gia Hân]

a, (2x-3)²=16

b, (3x-2)⁵=-243

c, (x-7)^x+1=(x-7)^x+11

Answer 1

[Lê Minh Vũ]

a) \(\left(2x-3\right)^2=16\)

\(\left(2x-3\right)^2=4^2\)

\(2x-3=4\)

\(2x=7\)

\(x=\dfrac{7}{2}=3,5\)

b) \(\left(3x-2\right)^5=-243\)

\(\left(3x-2\right)^5=-3^5\)

\(3x-2=-3\)

\(3x=-1\)

\(3x=-\dfrac{1}{3}\)

c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)

\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)

\(x-7=0;\left(x-7\right)^{10}=1\)

\(x=7;\left(x-7=1;x-7=-1\right)\)

\(x=7;x=8;x=6\)

Answer 2

[Nguyễn Thị Thương Hoài]

a, (2\(x\) - 3)² = 16

     \(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}

b, (3\(x\) - 2)⁵   = -243 

    ( 3\(x\) - 2)⁵ =  (-3)⁵

 3\(x\)    -  2     = -3

     3 \(x\)           = -1

        \(x\)          = - \(\dfrac{1}{3}\)

Vậy \(x\) = -\(\dfrac{1}{3}\)

c, \(\left(x-7\right)\)^\(x+1\)  = (\(x-7\))^\(x+11\)

    (\(x-7\))^{\(^{x+1}\)}.( \(\left(x-7\right)^{10}\) -  1 ) = 0

      \(\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)

         ^{\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)}

         \(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)

Vậy \(x\in\){ 6; 7; 8}