(x2 + 1)x = 0 => x2 + 1 = 0 hoặc x = 0
Mà: x2 \(\ge\)0 => x2 + 1\(\ge\)1 > 0 => Trường hợp x2 + 1 = 0 loại
=> x = 0
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Các câu hỏi tương tự
Bài 1 : tìm x biết
a) ( /x/ - 1/4 ) . ( x2 - 9 ) = 0
b) ( /x/ + 2 ) . ( /x/ - 4 ) = 0
c) ( x2 - 1/4 ) . ( x2 - 1/10 ) = 0
d) ( x + 2 ) . ( x - 3 ) < 0
e) ( x - 1/4 ) . ( x + 1/2 ) > 0
Bài 1 : Tìm x biết
a) ( /x/ - 1/4 ) . ( x2 - 9 ) = 0
b) ( /x/ + 2 ) . ( /x/ - 4 ) = 0
c) ( x2 - 1/4 ) . ( x2 - 1/16 ) = 0
d) ( x + 2 ) . ( x - 3 ) < 0
e) ( x - 1/4 ) . ( x + 1/2 ) > 0
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|x-3| + |x-2y+1| =0
|x-2| + |3x .y-5| =0
|x-2| - 3|2-x| =-10
|x-3|+ |x-y+1|=0
|x-1| + (y+5)2020 =0
tìm x
(x-1).(x+3)<0
(2-1).(x+5) >0
(x-5).(x+1/2)>0
(x+1).(x-3/2)<0
Tìm x biết : x mũ 2 + ( x - 1 ) mũ 2 = 0 , ( x -1 ) . ( x - 5 ) > 0 , ( x + 1 ) . ( x - 2 ) < 0
Tìm x,y biết :
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|x+1|-|x-2|=0
|x+2|-|x-3|=0
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Tìm x, x thuộc Q:
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b) (x - 2) * (x + 2/3) > 0
c) x + 1/2 * (x - 1/3) < 0
Bt1 tìm x
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a) 5/2 - x + 4/5 = 2/3 + 4/7
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