c, | x + 1| + | x + 2 | + | x + 3 | = 4x (1)
Ta có \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\forall x\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+3>x+2>x+1>x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\\left|x+3\right|=x+3\end{matrix}\right.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=x+1+x+2+x+3=3x+6\) (2)
Từ (1) và (2) \(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow x=6\)
Vậy x = 6
Cách khác của câu b).
b) \(5^{-1}.25^x=125\left(x\in Z\right).\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
c, vì /x+1/+/x+2/+/x+3/ ≥ 0
suy ra: 4x≥0 suy ra: x≥
suy ra :x+1+x+2+x+3=4x
suy ra x=6
vậy x=6
b, =5^-1.5^2x=5^3
5^-1+2x=5^3
suy ra:-1+2x=3
x=2
a, \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}\)
A)giả sử a>1 thì a2 <a4 => x-1>1 vô lý
giả sử a<(-1 ) thì a2<a4=>x-1< (-1) vô lý
=> a-1 =( -1;0;1)=> a =(1;0;2)