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Question

1)

$\sqrt{2x+1}+\sqrt{3-2x}+4+2\sqrt{3+4x-4x^2}=\frac{1}{4}(4x^2-4x+3)(2x-1)^2$

2)

$x(4x^2+1)+(x-3)\sqrt{5-2x}=0$

@_@ *Mèo9119* @_@ · Answer 1 · 4/19/2016 6:10:50 PM

TXĐ :

$x \in \left[ \frac{-1}2 ; \frac 32\right]$

$VT=[\sqrt{2x+1}+\sqrt{3-2x}]+[(\sqrt{2x+1})^2+2\sqrt{(2x+1)(3-2x)}+(\sqrt{3-2x})^2]$

$=(\sqrt{2x+1}+\sqrt{3-2x})+(\sqrt{2x+1}+\sqrt{3-2x})^2$

~~~~

$VP=\frac 14(4x^2-4x+3)(4x^2-4x+1)=\frac 14 \left[ (4x^2-4x+1)^2+2(4x^2-4x+1) \right]$

$=\left( \frac{4x^2-4x+1}{2} \right)^2+\left( \frac{4x^2-4x+1}{2} \right)$

~~~

Xét

$f(t)=t^2+t$ trên

$[0;+\infty)$ có

$f'(t)=2t \ge0 \forall t \in [0;+\infty)$

Nên

$f(t)$ là hàm đồng biến trên

$(0;+\infty)$

Ta có

$VT=VP\Leftrightarrow f\left( \sqrt{2x+1}+\sqrt{3-2x} \right )=f \left(\frac{4x^2-4x+1}2 \right)$

$\Leftrightarrow \sqrt{2x+1}+\sqrt{3-2x}=\frac{4x^2-4x+1}2$

Tới đây e bí :)))

tran85295 · Answer 2 · 4/18/2016 9:13:31 PM

Trả lời 18-04-16 09:13 PM

tran85295
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