Đáp án:
Giải thích các bước giải:
`h) (x-1)^2=(x-1)^4`
`=> (x-1)^2-(x-1)^4=0`
`=> (x-1)^2-(x-1)^2. (x-1)^2=0`
`=> (x-1)^2.[1-(x-1)^2]=0`
`=>` \(\left[ \begin{array}{l}(x-1)^2=0\\1-(x-1)^2=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=1\\(x-1)^2=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=1\\x-1=1\\x-1=-1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.\)
`h) 5^(-1) . 25^x = 125`
`=> 1/5 . 25^x=125`
`=> 25^x=625`
`=> 25^x=25^2`
`=> x=2`
`k) |x+1|+|x+2|+|x+3|=4x`
`TH_1: x < 0`
`=> 4x<0`
Mà `|x+1|+|x+2|+|x+3|>=0 ∀ x`
`=> |x+1|+|x+2|+|x+3|=4x` (vô lí)
`TH_2: x>0`
`=> (x+1)+(x+2)+(x+3)=4x`
`=> x+x+x-4x=-1-2-3`
`=> -x=-6`
`=> x=6 (TMĐK)`
Vậy `x=6`
