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Question

$\frac{1+3\sqrt{x}}{4x+2\sqrt{x+2}} - 1 =0$

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

$\sqrt[3]{x^{2}+4} = \sqrt{x-1} +2x-3$

tran85295 · Answer 1 · 1/31/2016 10:36:52 AM

Đk $x \ge 1$

$pt \Leftrightarrow(x-\sqrt[3]{x^2-4})+(x-2)+(\sqrt{x-1}-1)=0$

$\Leftrightarrow \frac{x^3-x^2-4}{x^3+x\sqrt[3]{x^2-4}+\sqrt[3]{(x^2+4)^2}}+(x-2)+\frac{x-2}{\sqrt{x-1}+1}=0$

$\Leftrightarrow (x-2)(\frac{x^2+x+2}{x^2+x\sqrt[3]{x^2-4}+\sqrt[3]{(x^2+4)^2}}+1+\frac1 {\sqrt{x-1}+1})=0$

Ngoặc thứ 2 $>0 \Rightarrow x=2$

Trả lời 31-01-16 10:36 AM

tran85295
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tran85295 · Answer 2 · 1/31/2016 10:30:12 AM

Đk $3x^3-2 \ge0 \Leftrightarrow x \ge \sqrt[3]{\frac 23}$

$pt \Leftrightarrow\sqrt[3]{x^2-1}+\sqrt{3x^3-2}-(3x-2)=0$

$\Leftrightarrow \sqrt[3]{x^2-1}+\frac{3x^3-2-(3x-2)^2}{\sqrt[3]{x^3-2}+3x-2}=0$

$\Leftrightarrow \sqrt[3]{x-1}[\sqrt[3]{x+1}+\frac{\sqrt[3]{(x-1)^2}(x^2-2x+2)}{\sqrt[3]{x^3-2}+3x-2}]=0$

Từ đkxđ dễ thấy thừa số thứ 2 $>0$ $\Rightarrow x=1$

Trả lời 31-01-16 10:30 AM

tran85295
141 5

10M 559K

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