(3x-7)^2018 = (3x-7)^2016
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24 tháng 7 2019
a) \(\left(4x^2-2\right)^2=\frac{196}{81}\)
<=> \(2^2\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(4\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(\left(2x^2-1\right)^2=\frac{196}{81}:4\)
<=> \(\left(2x^2-1\right)^2=\frac{49}{81}\)
<=> \(2x^2-1=\pm\sqrt{\frac{49}{81}}\)
<=> \(2x^2-1=\pm\frac{7}{9}\)
<=> \(\orbr{\begin{cases}2x^2-1=\frac{7}{9}\\2x^2-1=-\frac{7}{9}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
CW
7 tháng 7 2016
1) x (x-2016) + 2015 (2016-x) = 0
x (x-2016) - 2015 (x- 2016) = 0
(x-2015)(x-2016) =0
\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)
Vậy x= 2015; 2016
2) -5x (x-15) + (15-x) = 0
-5x (x-15) - (x-15) =0
(-5x -1) (x-15) =0
\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)
Vậy x= -1/5; 15
3) 3x (3x-7) - (7-3x) =0
3x(3x-7) + (3x -7) =0
(3x+1) (3x-7) =0
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)
Vậy x= -1/3 ; 7/3
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KK
5 tháng 9 2021
2016 : \(\left[25-\left(3x+2\right)\right]=3^7\)
<=> 25 - (3x + 2) = 2016 : 2187
<=> 25 - 3x - 2 = \(\dfrac{224}{243}\)
<=> -3x = \(\dfrac{224}{243}\) + 2 - 25
<=> -3x = \(\dfrac{-5365}{243}\)
<=> x = 7,359396433 \(\approx\) 7,4
\(\left(3x-7\right)^{2018}=\left(3x-7\right)^{2016}\)
\(\Leftrightarrow\left(3x-7\right)^{2018}-\left(3x-7\right)^{2016}=0\)
\(\Leftrightarrow\left(3x-7\right)^{2016}\left[\left(3x-7\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-7\right)^{2016}=0\\\left(3x-7\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(3x-7\right)^{2016}=0\\\left(3x-7\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\3x-7=1\\3x-7=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x=7\\3x=8\\3x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{8}{3}\\x=2\end{matrix}\right.\)