[Toán 8] Đề HSG<img src="http://diendan.hocmai.vn/images/eyeeasy/buttons/ChuaXN.png" border="0">

K

kisihoangtoc

1a

$(a-1)^2$ \geq $0$\Rightarrow$a^2+1$ \geq $2a$
$(b-1)^2$ \geq $0$\Rightarrow$b^2+1 $\geq $2b$
$(a-b)^2$ \geq $0$\Rightarrow$a^2+b^2$ \geq $2ab$
\Rightarrow $2(a^2+b^2+1)$ \geq $2(a+b+ab)$
\Rightarrow $a^2+b^2+1$ \geq $a+b+ab$
 
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Z

z0987654321

bài 1:
a)=>$2a^2+2b^2+2$\geq $2ab+2a+2b$
=>[TEX](a-b)^2+(a-1)^2+(b-1)^2\geq0[/TEX] luôn đúng => DPCM
b) =>[TEX]a^3(a-b)-b^3(a-b)\geq0[/TEX]=>[TEX](a-b)(a^3-b^3) \geq 0[/TEX]
=>[TEX](a-b)^2[(a+\frac{b}{2})^2+\frac{b^2}{3}]\geq0[/TEX] luôn đúng => DPCM
2) gt => [TEX]2x^2+2y^2-2xy=8[/TEX]=> [TEX]x^2+y^2+(x-y)^2=8[/TEX]
=>[TEX]x^2+y^2 \leq 8 [/TEX] dáu = xảy ra khi x=y và them gt
gt=>[TEX]2x^2+2y^2-2xy=8[/TEX]=>[TEX]3x^2+3y^2=8+(x-y)^2[/TEX]\
=> [TEX]x^2+y^2\geq\frac{8}{3}[/TEX] dấu = xảy ra khi x=-y và thêm gt
 
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K

kisihoangtoc

1b


$a^4 + b^4$ \geq $a^3b + ab^3$
\Leftrightarrow $a^4-a^3b+b^4-ab^3$ \geq $0$
\Leftrightarrow $a^3(a-b)-b^3(a-b)$ \geq $0$
\Leftrightarrow $(a^3-b^3)(a-b)$ \geq $0$
\Leftrightarrow $(a-b)^2(a^2+ab+b^2)$ \geq $0$(luôn đúng)
 
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K

kisihoangtoc

2


$(x-y)^2$ \geq $0$ \Rightarrow $\frac{x^2+y^2}{2}$ \geq $xy$ \Rightarrow $\frac{-(x^2+y^2)}{2}$ \leq $-xy$
$4=x^2+y^2-xy$ \geq $x^2+y^2 - \frac{x^2+y^2}{2} = \frac{x^2+y^2}{2}$
\Rightarrow $x^2+y^2$ \leq $8$
Vậy $Max A = 8$
$(x+y)^2$ \geq $0$ \Rightarrow $\frac{x^2+y^2}{2}$ \geq $-xy$
$4=x^2+y^2-xy$ \leq $x^2+y^2 + \frac{x^2+y^2}{2} = \frac{3(x^2+y^2)}{2}$
\Rightarrow $x^2+y^2$ \geq $\frac{8}{3}$
Vậy $Min A = \frac{8}{3}$
 
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