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\(\left(8x-1\right)^{16}=\left(8x-1\right)^{18}\)
\(\Leftrightarrow\left(8x-1\right)^{18}-\left(8x-1\right)^{16}=0\)
\(\Leftrightarrow\left(8x-1\right)^{16}\left[\left(8x-1\right)^4-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(8x-1\right)^{16}=0\\\left[\left(8x-1\right)^4-1\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}8x-1=0\\\left(8x-1\right)^4-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}8x-1=0\\\left[{}\begin{matrix}8x-1=1\\8x-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=1\\\left[{}\begin{matrix}8x=2\\8x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
\(\left(x^2-8x\right)-3x+24=0\)
\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-8\right)=0\Leftrightarrow x=3;x=8\)
bài 59; tìm x biết
2, ( x mũ 2 - 8x ) - 3x + 24 = 0
x= 3 ; x = 8
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
*\(x-2=1\Rightarrow x=3\)
*\(x-2=-1\Rightarrow x=1\)
Vậy x = 3; x = 1
c) \(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vậy x = \(\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(\Rightarrow x=\frac{-1}{4}\)
Vậy x = \(\frac{-1}{4}\)
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=\left(\pm1\right)^2\)
\(\begin{cases}x-2=1\\x-2=-1\end{cases}\)
\(\begin{cases}x=1+2\\x=-1+2\end{cases}\)
\(\begin{cases}x=3\\x=1\end{cases}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\)
\(\begin{cases}x=\frac{1}{4}-\frac{1}{2}\\x=-\frac{1}{4}-\frac{1}{2}\end{cases}\)
\(\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)
\(\left(\frac{1}{4}\right)^x=\frac{1}{4}\cdot\frac{1}{16}\)
\(\left(\frac{1}{4}\right)^x=\frac{1}{64}\)
\(\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)
\(\Rightarrow x=3\)
b) \(81^x:9^x=729\)
\(\left(81:9\right)^x=729\)
\(9^x=9^3\)
\(\Rightarrow x=3\)
a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{4}\times\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{64}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)
\(\Rightarrow x=3\)
Vậy x = 3
b) \(81^x:9^x=729\)
\(\Rightarrow\left(81:9\right)^x=729\)
\(\Rightarrow9^x=9^3\)
\(\Rightarrow x=3\)
Vậy x = 3
_Chúc bạn học tốt_
8x=162
8x=256
x=256:8
x=32
nha
\(8x=16^2=256\)
\(x=256:8=32\)
Mình nghĩ là \(16^2\) chứ không phải \(16^{12}\) nha!