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3 tháng 9 2019
a. \(5^{4-x}+1=26\)
\(\Leftrightarrow5^{4-x}=26-1=25\)
\(\Leftrightarrow5^{4-x}=5^2\)
\(\Leftrightarrow4-x=2\)
\(\Leftrightarrow x=2\)
b. \(\left(\frac{2}{x}+1\right)^{2x}=5^{2x}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}+1=5\\\frac{2}{x}+1=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=4\\\frac{2}{x}=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{3}\end{cases}}\)
c. \(\left(1-2x\right)^4-\left(1-2x\right)^6=0\)
\(\Leftrightarrow\left(1-2x\right)^4.\left[1-\left(1-2x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(1-2x\right)^4=0\\1-\left(1-2x\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}1-2x=0\\\left(1-2x\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=1\\2x=0hoac2x=-2\end{cases}}\)
\(\Leftrightarrow x=\frac{1}{2},x=0,x=-1\)
NH
1
TM
2
28 tháng 7 2019
a) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)
\(\Leftrightarrow\left(2.x-1\right)^8-\left(2.x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},1\right\}\)
28 tháng 7 2019
b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy : \(x\in\left\{0,1,2\right\}\)
Chúc học tốt nhé !!
\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
Đặt 2x + 1 = a, ta có
\(a^4=a^6\)
\(\Rightarrow a^4-a^6=0\)
\(\Rightarrow a^4\left(1-a^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a^4=0\\1-a^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\\2x+1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\2x=0\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)
\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Rightarrow\left(2x+1\right)^4-\left(2x+1\right)^6=0\)
\(\Rightarrow\left(2x+1\right)^4.\left[\left(2x+1\right)^2-1\right]0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left[\left(2x+1\right)^2-1\right]=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=0\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=\dfrac{-1}{2};x_2=0\right\}\)