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Cho a+x2=2006, b+x2=2007, c+x2= 2008 và abc=3
Tính a/bc+b/ca+c/ab-1/a-1/b-1/c
.
a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)
Ta có : a2 + b2 + c2 = ab + bc + ca
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
= (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0
=> (a - b)2 + (b - c)2 + (c - a)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
b) Ta có : 2(x2 + t2) + (y + t)(y - t) = 2x(y + t)
=> 2x2 + 2t2 + y2 - t2 = 2xy + 2t
=> 2x2 + t2 + y2 = 2xt + 2xy
=> 2x2 + t2 + y2 - 2xt - 2xy = 0
=> (x2 - 2xy + y2) + (x2 + t2 - 2xt) = 0
=> (x - y)2 + (x - t)2 = 0
=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)
c) Ta có a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> a2 + b2 + c2 = 0
=> a = b = c = 0
Khi đó A = (0 - 1)2003 + 02004 + (0 + 1)2005
= - 1 + 0 + 1 = 0
Vậy A = 0
1)
$2x=a+b+c$
$\Rightarrow x-a=\dfrac{b+c-a}{2},\quad x-b=\dfrac{c+a-b}{2},\quad x-c=\dfrac{a+b-c}{2}$
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$
$=\dfrac{(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)}{4}$
$=\dfrac{2ab+2bc-a^2-b^2-c^2+2bc+2ca-a^2-b^2-c^2+2ca+2ab-a^2-b^2-c^2}{4}$
$=\dfrac{4ab+4bc+4ca-3(a^2+b^2+c^2)}{4}$
$=\dfrac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$
$=\dfrac{(2x)^2-2(a^2+b^2+c^2)}{4}$
$=ab+bc+ca-x^2$
$\therefore\ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+ac+bc-x^2.$
2)
$ab+bc+ca=abc,\quad a+b+c=1$
$(a-1)(b-1)(c-1)$$=abc-ab-bc-ca+a+b+c-1$$=abc-ab-bc-ca+1-1$$=abc-(ab+bc+ca)$$=abc-abc$$=0$
$\therefore\ (a-1)(b-1)(c-1)=0.$
3)
$x-y=12$
$A=x^3-y^3-36xy$
$=(x-y)(x^2+xy+y^2)-36xy$
$=12(x^2+xy+y^2)-36xy$
$=12(x^2-2xy+y^2)+36xy$
$=12(x-y)^2+36xy$
$=12\cdot12^2$
$=1728$
$\boxed{A=1728}$