Bài 27:
$x^2+5y^2+2xy-4x-8y+2015$
$=x^2+2xy+y^2-4x-4y+4+4y^2-4y+1+2010$
$=(x+y)^2-4(x+y)+4+(2y-1)^2+2010$
$=(x+y-2)^2+(2y-1)^2+2010\geq 2010$
Dấu "=" xảy ra khi $x=\dfrac{3}{2};y=\dfrac{1}{2}$
Vậy...
Bài 28:
$a+b+c=0\Rightarrow c=-(a+b);a=-(b+c);b=-(c+a)$
$\Rightarrow a^2+b^2-c^2=-2ab;b^2+c^2-a^2=-2bc;c^2+a^2-b^2=-2ca$
Vậy $P=\dfrac{1}{-2ab}+\dfrac{1}{-2bc}+\dfrac{1}{-2ca}=\dfrac{z+x+y}{-2xyz}=0$
Bài 29:
$a^3+b^3+c^3=3abc$
$\iff a^3+b^3+c^3-3abc=0$
$\iff (a+b)^3+c^3-3ab(a+b+c)=0$
$\iff (a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0$
$\iff (a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0$
$\iff 2(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0$
$\iff (a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]=0$
+Nếu $a+b+c=0$ thì $P=\dfrac{a+b}{b}+\dfrac{b+c}{c}+\dfrac{c+a}{a}=\dfrac{-c}{b}+\dfrac{-a}{c}+\dfrac{-b}{a}=-1$
+Nếu $(a-b)^2+(b-c)^2+(c-a)^2=0\Rightarrow a=b=c$ thì $P=(1+1)(1+1)(1+1)=8$
Bài 32:
Áp dụng k/q ở bài 29 nếu $x+y+z=0$ thì $x^3+y^3+z^3=3xyz$ ta đc:
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow \dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})=\dfrac{3}{abc}$
mà $\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}$
=> $A=abc.\dfrac{3}{abc}=3$
Bài 33:
$x^3+ax+b$ hay $x^2+ax+b$ vậy?
Bài 34:
Thay $2017=abc$ ta đc $C=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{ac+abc^2+abc}\\=\dfrac{1}{b+1+bc}+\dfrac{b}{bc+b+1}+\dfrac{bc}{1+bc+b}=1$
[TEX]a^3+b^3+c^3=3abc[/TEX]
[TEX](a+b)^3-3ab(a+b)+c^3-3abc=0[/TEX]
[TEX][(a+b)^3+c^3]-3ab(a+b+c)=0[/TEX]
[TEX](a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0[/TEX]
rồi đặt nhân tử chung a+b+c ra ngoài
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