Vì:  |2\(x\) - 1| = |1 - 2\(x\)|

Nên:      |2\(x\) - 1| + |1 - 2\(x\)| = 8

  ⇒        |2\(x\) - 1| + |2\(x\) - 1| = 8

                         2.|2\(x\) - 1| = 8

                            |2\(x\) - 1| = 8:2

                           |2\(x\) - 1| = 4

                         \(\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\)

                          \(\left[{}\begin{matrix}2x=-4+1\\2x=4+1\end{matrix}\right.\)

                            \(\left[{}\begin{matrix}2x=-3\\2x=5\end{matrix}\right.\)

                             \(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\){- \(\dfrac{3}{2}\); \(\dfrac{5}{2}\)}

giúp mình với

Ta có: \(\left\{{}\begin{matrix}\left(2x-3y\right)^{2018}\ge0\forall x,y\\\left(3y-4z\right)^{2020}\ge0\forall y,z\\\left|2x+3y-z-63\right|\ge0\forall x,y,z\end{matrix}\right.\)

\(\Rightarrow\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|\ge0\forall x,y,z\)

Mà: \(\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|=0\)

nên: \(\left\{{}\begin{matrix}2x-3y=0\\3y-4z=0\\2x+3y-z-63=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x=3y\\3y=4z\\z=2x+3y-63\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=4z\\3y=4z\\z=4z+4z-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4z:2\\y=4z:3\\z=8z-63\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=4z:3\\-7z=-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=4\cdot9:3=12\\z=9\end{matrix}\right.\)

Vậy \(x=18;y=12;z=9\).

$Toru$

đề b là j vậy

tui cần bài đó thôi

1,Thay x = 1 vào biểu thức ta có

f = 4 x 1² -5

f = -1

2, Đặt f(x) = -1, ta có:

4 x x² - 5 = -1

4 x x² = 4

x² = 4 : 4

x² = 1

x²=1²

=> x = 1 hoặc = -1

Vậy để f(x)=1 thì x ϵ {-1;1}