Do \(\left|10-x\right|,\left|8-x\right|\ge0\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}10-x=0\\8-x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=10\\x=8\end{matrix}\right.\)(Vô lý)
Vậy \(S=\varnothing\)
Tìm x biết | 10-x| + | 8-x | =0
Tìm x biết
a) (x-8)(x^3+8)=0
b) (4x-3)-(x+5)=3(10-x)
Tìm x, biết:
x^10-x^8=0
tìm x biết
a/ (x-8)(x^3 +8)=0
b/ (4x-3)-(x+5)=3(10-x)
Tìm x biết :
a) ( 2x - 3 ).( x +1 ) > 0
b) ( x + 5 ).(x-7) < 0
c) | 2x - 3 | + 8 = 10
d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
Tìm x : \(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)0
tìm x, biết:
a, (x - 8) (x3 + 8) = 0
b, (4x - 3) - (x + 5) = 3(10-x)
10, Tìm x.
a) |x-2|+|x-8|=6
b) |2x-1|+|y-3x|=0
