Question

[Lê Thị Cẩm Ly]

^{(x-3)^2 =16}

^{(1-3x )^3 =-64}

^{x^13 = 27. x^10}

^{(4x-1)^2 = (1 -4x )^4}

Answer 1

[Nguyễn Thanh Hiền]

+) \(\left(x-3\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=4^2\\\left(x-3\right)^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

Vậy x = 7 hoặc x = -1

+) \(\left(1-3x\right)^3=-64\)

\(\Rightarrow\left(1-3x\right)^3=\left(-4\right)^3\)

\(\Rightarrow1-3x=-4\)

\(\Rightarrow3x=1+4\)

\(\Rightarrow3x=5\)

\(\Rightarrow x=5:3\)

\(\Rightarrow x=\frac{5}{3}\)

Vậy  \(x=\frac{5}{3}\)

+) \(x^{13}=27.x^{10}\)

\(\Rightarrow x^{13}:x^{10}=27\)

\(\Rightarrow x^3=27\)

\(\Rightarrow x^3=3^3\)

\(\Rightarrow x=3\)

Vậy x = 3

+) \(\left(4x-1\right)^2=\left(1-4x\right)^4\)

\(\Rightarrow\left(4x-1\right)^2=\left(4x-1\right)^4\)

\(\Rightarrow\left(4x-1\right)^2-\left(4x-1\right)^4=0\)

\(\Rightarrow\left(4x-1\right)^2\left[1-\left(4x-1\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\\left(4x-1\right)^2=1\end{cases}}\)

TH 1 : \(\left(4x-1\right)^2=0\Rightarrow4x-1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\)

TH 2 : \(\left(4x-1\right)^2=1\Rightarrow\orbr{\begin{cases}4x-1=1\\4x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}4x=2\\4x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)

Vậy  \(x\in\left\{\frac{1}{4};\frac{1}{2};0\right\}\)

_Chúc bạn học tốt_

Answer 2

[Trần Thanh Thảo]

a, (x-3)^2 = 16

=> (x-3)^2=4^2

=> x-3=4

=> x= 4+3

=> x = 7 .Vậy x =7

b,(1-3x)^3 = 64

=> ( 1-3x)^3 = 4^3

=> 1-3x = 4

=> 3x = 1-4

=> 3x = -3

=> x = -1 . Vậy x = -1

c, x^13 = 27.x^10

=> x^13 : x^10 = 27

=> x^3 = 3^3

=> x = 3 . Vậy x = 3