Toán 8

N

nice_simple

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N

nguyenbahiep1

câu 1

[laTEX]A = x^2 + xy + y^2 -3x - 3y + 3 \\ \\ A = x^2+x(y-3) +\frac{y^2-6y+9}{4} + y^2 -3y + 3 - \frac{y^2-6y+9}{4} \\ \\ A = (x+ \frac{y-3}{2})^2 + \frac{3}{4}.(y-1)^2 \geq 0 \Rightarrow dpcm[/laTEX]
 
H

huytrandinh

câu 1
[TEX]<=>(x+y)^{2}-3(x+y)-xy+3\geq 0 (1)[/TEX]
[TEX].(x-y)^{2}\geq 0<=>(x+y)^{2}\geq 4xy[/TEX]
[TEX]=>-xy\geq -\frac{1}{4}(x+y)^{2}[/TEX]
[TEX]=>VT(1)\geq \frac{3}{4}(x+y)^{2}-3(x+y)+3\geq 0[/TEX]
[TEX]<=>(x+y-2)^{2}\geq 0[/TEX]
câu 2
[TEX]<=>x^{3}(x+y)+y^{3}(x+y)\geq 0[/TEX]
[TEX]<=>(x+y)(x^{3}+y^{3})\geq 0[/TEX]
[TEX]<=>(x+y)^{2}(x^{2}-xy+y^{2})\geq 0[/TEX]
[TEX]<=>(x+y)^{2}[(x+\frac{1}{2}y)^{2}+\frac{3}{4}y^{2}]\geq 0[/TEX]
câu 3
bđt sai ví dụ (a,b,c) là (0.5,0.4,0.2)
câu 4
[TEX].(|a|-|b|)^{2}\geq 0<=>a^{2}+b^{2}\geq 2|ab|[/TEX]
[TEX].(|a|-1)^{2}\geq 0[/TEX]
[TEX]<=>a^{2}+1\geq 2|a|[/TEX]
[TEX]=>(a^{2}+b^{2})(a^{2}+1)\geq 2|ab|.2|a|=4a^{2}|b|[/TEX]
[TEX]\geq 4a^{2}b[/TEX]
 
N

noinhobinhyen

C.5 sai đề

(a;b;c;d)=(1;10;10;1) ko thỏa mãn
-------------------------------------------

C.3 ko sai

$a^4+b^4 \geq 2a^2b^2$

$b^4+c^4 \geq 2b^2c^2$

$a^4+c^4 \geq 2a^2c^2$

Cộng lại ...
 
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