4 replies to this topic

[ngocdz9apro]

1.$(x^{2}-6x+11)\sqrt{x^{2}-x+1}=2(x^{2}-4x+7)\sqrt{x-2}$

2.$\sqrt{x^{2}+7}=4x-10+\sqrt{x^{2}-5}$

3.$(x-x^{2})(x^{2}+3x+2013)-2011x\sqrt{4-4x}=30\sqrt[4]{x^{2}+x-1}+2012$

4.$(3-3)\sqrt{x-1}+\sqrt{5-2x}=\sqrt{-x^{3}+10x^{2}-34x+40}$

5.$\sqrt{x^{2}-8x+816}+\sqrt{x^{2}+10x+267}=\sqrt{2003}$

 

[sharker]

Bài 5

$\sqrt {{x^2} - 8x + 816}  + \sqrt {{x^2} + 10x + 267}  = \sqrt {2003} $

$VT = \sqrt {{{(4 - x)}^2} + 800}  + \sqrt {{{(x + 5)}^2} + 242}  \ge \sqrt {{{(4 - x + x + 5)}^2} + {{(\sqrt {800}  + \sqrt {242} )}^2}}  = 2003$

Dấu = xảy ra khi $x=\frac{-56}{31}$

Edited by sharker, 18-04-2017 - 12:35.

[sharker]

1.$(x^{2}-6x+11)\sqrt{x^{2}-x+1}=2(x^{2}-4x+7)\sqrt{x-2}$

 

Bình phương 2 vế ta được :$x^6-17x^5+111x^4-386x^3+775x^2-897x+513=0$

$\Leftrightarrow (19 - 10 x + x^2) (9 - 5 x + x^2) (3 - 2 x + x^2)=0$

[HoangKhanh2002]

2.$\sqrt{x^{2}+7}=4x-10+\sqrt{x^{2}-5}$

Làm tạm, chiều gửi hết

ĐK: $-\sqrt{5}\leq x\leq \sqrt{5}$

$\sqrt{x^{2}+7}=4x-10+\sqrt{x^{2}-5}\Leftrightarrow \sqrt{x^2+7}-4=4x-12+\sqrt{x^2-5}-2\Leftrightarrow \frac{x^2-9}{\sqrt{x^2+7}+4}=4(x-3)+\frac{x^2-9}{\sqrt{x^2-5}+2}\Leftrightarrow (x-3)\left ( \frac{x+3}{\sqrt{x^2+7}+4}-4-\frac{x+3}{\sqrt{x^2-5}+2} \right )=0\Rightarrow x=3$

Vì $\frac{x+3}{\sqrt{x^2+7}+4}-4-\frac{x+3}{\sqrt{x^2-5}+2}<-4<0$

[ngocdz9apro]

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