Question

[Lê Hoàng Tiến Đạt]

a) \(-5+|3x-1|+6=|-4|\)

b)\(\left(x-1\right)^2=\left(x-1\right)^4\)

c)\(5^{-1}.25^x=125\left(x\in Z\right)\)

d)\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)

Answer 1

[Kuroba Kaito]

a) -5 + |3x - 1| + 6 = |-4|

=> -5 + |3x - 1| + 6 = 4

=> 1 + |3x - 1| = 4

=> |3x - 1| = 4 - 1

=> |3x - 1| = 3

=> \(\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}}\)

=> \(\orbr{\begin{cases}3x=4\\3x=-2\end{cases}}\)

=>  \(\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy ...

d) |x + 1| + |x + 2| + |x + 3| = 4x

Ta có: |x + 1| \(\ge\)0  \(\forall\)x

          |x + 2| \(\ge\)0 \(\forall\)x

          |x + 3| \(\ge\)0 \(\forall\)x

=> |x + 1| + |x + 2| + |x + 3| \(\ge\)0 \(\forall\)x => 4x \(\ge\)0 \(\forall\) x=> x \(\ge\)0 \(\forall\)x

=> x + 1 + x + 2 + x + 3 = 4x

=> 3x + 6 = 4x

=> 6 = 4x - 3x

=> x = 6

Vậy...

Answer 2

[Kuroba Kaito]

b) (x - 1)² = (x - 1)⁴

=> (x - 1)² - (x - 1)⁴ = 0

=> (x - 1)² .[1 - (x - 1)² ] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\)

=> \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy x = {1; 2; 0}