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a) Xem lại đề
b) \(\Leftrightarrow3x^4-2x^2-1=0\Leftrightarrow\left(3x^2+1\right)\left(x^2-1\right)=0\Leftrightarrow x^2=1\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a) \(\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4\)
\(\Leftrightarrow\frac{1}{3^{4x}}\cdot3^{6x}=9^4\)
\(\Leftrightarrow\frac{3^{6x}}{3^{4x}}=3^8\)
\(\Leftrightarrow3^{2x}=3^8\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
b) \(5^x\cdot\left(5^3\right)^2=625\)
\(\Leftrightarrow5^{x+6}=5^4\)
\(\Leftrightarrow x+6=4\)
\(\Leftrightarrow x=-2\)
c) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^{10}=1=\left(\pm1\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{1}{2}\\x=0\end{matrix}\right.\)
Vậy....
a) 4x - 15 = - 75 - x
<=> 4x + x = -75 + 15
<=> 5x = -60
<=> x = -12
b) 72 - 3x = 5x + 8
<=> -3x -5x = 8 - 72
<=> -8x = -64
<=> x = 8
d) 3 . | x - 7 | = 21
<=> | x - 7 | = 7
\(\Leftrightarrow\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=0\end{cases}}\)
\(-7\left|x+3\right|=-49\)
\(\Leftrightarrow\left|x+3\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)
a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)
=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)
=> \(2x+\frac{3}{4}=-\frac{7}{4}\)
=> \(2x=\frac{-7}{4}-\frac{3}{4}\)
=> \(2x=-\frac{5}{2}\)
=> \(x=\frac{-5}{2}:2\)
=> \(x=\frac{-5}{4}\)
b) \(\frac{x+1}{3}=\frac{2-x}{2}\)
\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)
\(\Rightarrow2x+2=6-3x\)
\(\Rightarrow2x-3x=6-2\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=4\)
c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)
\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{1}{5}:\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-4x=0\)
Ta có : \(x^2-4x=0\)
\(\Rightarrow xx-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)