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william henry bill gates
william henry bill gates
Member Since 27-04-2016Offline Last Active 15-06-2019 - 17:11
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#723078 Hình học không giam
Posted by william henry bill gates on 15-06-2019 - 17:15
#643258 Chứng minh rằng: $(x^2-1)(y^2-1)(z^2-1)\le...
Posted by william henry bill gates on 02-07-2016 - 12:47
Đặt $ x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c} \rightarrow ab+bc+ca=1 và a,b,c<1 $
Bđt $\leftrightarrow (1-a^2)(1-b^2)(1-c^2) \leq abc\sqrt{(1+a^2)(1+b^2)(1+c^2)}=abc(a+b)(b+c)(c+a) $Use:$(1-a^2)(1-b^2) \leq (1-ab)^2=c^2(a+b)^2 $
cộng lại suy ra đpcm
(y)
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#643256 Giải pt 1)$\sqrt[3]{4x+4}=\sqrt{3x+1}$
Posted by william henry bill gates on 02-07-2016 - 12:33
11)$2x^{2}+2x+1=\sqrt{4x+1}$
p$\dpi{100} <=> 4x^2+4x+2=2\sqrt{4x+1} <=> 4x^2+4x+2-2\sqrt{4x+1}=0 <=> 4x^2 +(\sqrt{(4x+1)-1})^{\2}=0 <=> x=0$
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#643255 Giải pt 1)$\sqrt[3]{4x+4}=\sqrt{3x+1}$
Posted by william henry bill gates on 02-07-2016 - 12:10
$5\sqrt{2(x+2)(x^2-2x+4)}=2(X^2+8) <=> 5\sqrt{(2x+4)(x^2-2x+4)}=2(x^2+8)..... \sqrt{2x+4}=a; \sqrt{x^2-2x+4}=b <=> a^2+b^2= x^2+8 .....5ab=2(a^2+b^2)<=>(a-2b)(b-2a)=0$
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