a) \(\left(4x^2-2\right)^2=\frac{196}{81}\)

<=> \(2^2\left(2x^2-1\right)^2=\frac{196}{81}\)

<=> \(4\left(2x^2-1\right)^2=\frac{196}{81}\)

<=> \(\left(2x^2-1\right)^2=\frac{196}{81}:4\)

<=> \(\left(2x^2-1\right)^2=\frac{49}{81}\)

<=> \(2x^2-1=\pm\sqrt{\frac{49}{81}}\)

<=> \(2x^2-1=\pm\frac{7}{9}\)

<=> \(\orbr{\begin{cases}2x^2-1=\frac{7}{9}\\2x^2-1=-\frac{7}{9}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)

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

ngu thế à bạn

mux 2 nha moi ng 

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

2016 : \(\left[25-\left(3x+2\right)\right]=3^2\)

<=> 25 - (3x + 2) = 2016 : 9

<=> 25 - 3x - 2 = 224

<=> -3x = 224 - 25 + 2

<=> -3x = 201

<=> x = \(\dfrac{201}{-3}=-67\)