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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Cho a+b+c=0,a^2+b^2+c^2=1.Tính a^4+b^4+c^4

Toán Lớp 8: Cho a+b+c=0,a^2+b^2+c^2=1.Tính a^4+b^4+c^4

Comments ( 1 )

  1. Ta có $a+b+c=0$
    -> $(a+b+c)^{2}=0$
    Hay $a^{2}+ b^{2}+c^{2} + 2(ab+bc+ca) = 0$ (hằng đẳng thức)
    hay $1 + 2(ab+bc+ca) = 0$
    -> $ab+bc+ca= -\frac{1}{2}$
    Đặt $ab=x; bc=y; ac=z$
    Ta có: $x+y+z= -\frac{1}{2}$
    -> $(x+y+z)^{2}=\frac{1}{4}$
    Hay$x^{2}+ y^{2}+z^{2} + 2(xy+yz+zx) = \frac{1}{4}$
    hay $(ab)^{2}+ (bc)^{2}+(ca)^{2} + 2abc(a+b+c) = \frac{1}{4}$
    ->$(ab)^{2}+ (bc)^{2}+(ca)^{2} = \frac{1}{4}$ (vì a+b+c=0) (a)
    Lại có $a^{2}+ b^{2}+c^{2}=1$
    Hay $(a^{2}+ b^{2}+c^{2})^{2}=1$
    Hay $(a^{4}+ b^{4}+c^{4}+2[(ab)^{2}+ (bc)^{2}+(ca)^{2}]=1$ (b)
    (a)(b)->$a^{4}+ b^{4}+c^{4} + \frac{1}{2}=1$
    ->$a^{4}+ b^{4}+c^{4}=\frac{1}{2}$

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222-9+11+12:2*14+14 = ? ( )