Toán khó nè 3` kon

H

healthyeconomy

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V

vansang02121998

Bài 1:
- Ta có [tex]x^2+x+1=0[/tex]
=> [tex]x^2+x=-1[/tex]
mà [tex]x^2 \geq x[/tex]
=> [tex]x^2+x \geq 0[/tex]
=> Không tìm được giá trị của x
=> A không có giá trị
 
Last edited by a moderator:
K

katoriitto

Bài 1:
- Ta có x^2+x+1=0
=> x^2+x=-1
=> x=x^2+1
mà x^2 \geq x
=> x^2+1 \geq x
=> Không tìm được giá trị của x
=> A không có giá trị
Nhớ cảm ơn mình nha
 
S

s2thiensutinhyeus2

Bài 1:
- Ta có x^2+x+1=0
=> x^2+x=-1
=> x=x^2+1
mà x^2 x
=> x^2+1 x
=> Không tìm được giá trị của x
=> A không có giá trị
Bài 2 đang nghĩ
 
C

changmongmo1903

Các bạn nói bài sai thì phải chỉ ra chỗ sai chứ, nói không đâu có được. Tớ nghĩ bài của bài của các bạn ấy giải đúng và rất logic. Không biết ý các bạn như thế nào, nhưng mình cùng giải y hệt vậy.
 
L

lamhaisonbd

bai 1 la dang bai tim nghiem cua da thuc thui ma. cac ban do giai dug chu dau co sai cho nao dau
 
H

hoa_giot_tuyet

Bài 1
Tính A(x)=x^n+1/(x^n)
Biết x^2+x+1=0
Bai2:
Chứng minh rằng:
1/(1.2)+1/(3.4)+1/(5.6)+...+1/(49.50)=1/26+1/27+1/28+...+1/50:confused::confused::confused:

Bài 1 đố mẹo đấy, dù nó không tồn tại nhưng bài vẫn làm đc :D
Có làm rồi

Trích lời sách
a thoả mãn [TEX]a^2+a+1=0[/TEX], nên một cách không tường minh người ta đã cho a trong trường các số phức C

Bài này hell_angel_1997 đăng lên một lần rồi :rolleyes:
Từ [TEX]n^2+n+1=0 \Rightarrow n \neq 1 \Rightarrow (n-1)(n^2+a+1) = 0 \Rightarrow a^3 - 1 = 0 \Rightarrow a^3 = 1[/TEX]
Xét 3 trường hợp:
_ VỚi n = 3k thì [TEX]A = (n^3)^k + \frac{1}{(n^3)^k} = 1 + 1 = 2 (n^3 = 1)[/TEX]
_ Với n = 3k + 1 thì [TEX]A = (n^3)^k.n + \frac{1}{(n^3)^k.n} = n + \frac{1}{n} = \frac{n^2+1}{n} = \frac{-n}{n} = -1[/TEX]
_Với n = 3k+2 thì [TEX]A = (n^3)^k.n^2 + \frac{1}{(n^3)^k.n^2} = n^2+\frac{1}{n^2}[/TEX]
Ta có [TEX](n+\frac{1}{n})^2 = n^2 + \frac{1}{n^2} + 2.n.\frac{1}{n} = n^2+\frac{1}{n^2} + 2 = 1[/TEX]
\Rightarrow A = 1 -2 = -1
Sai chỗ này chỉnh dùm ha :D
 
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