Chủ đề này có 4 trả lời

[nguyentrongtan15112000]

Bài1:

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

Bài2:

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

Bài 3

$\sqrt{x+2}=\frac{x+2+2\sqrt{2x+1}}{x+\sqrt{2x+1}}$

Bài 4

$x\sqrt{2x+3}+3(\sqrt{x+5}+1)=3x+\sqrt{2x^2+3x+15}+\sqrt{2x+3}$

Bài 5

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

[tungteng532000]

Bài1:

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

Bài 1: Casio
$PT\Leftrightarrow (2\sqrt{1-x}+\sqrt{x+1}-3)(\sqrt{1-x}+\sqrt{x+1}+2)=0\Leftrightarrow 2\sqrt{1-x}+\sqrt{x+1}-3=0$
Cái này dễ rồi   
 

[tungteng532000]

 

Bài2:

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

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

[tungteng532000]

 

Bài 3

$\sqrt{x+2}=\frac{x+2+2\sqrt{2x+1}}{x+\sqrt{2x+1}}$

$PT\Leftrightarrow (\sqrt{x+2}-2)+(2-\frac{x+2+2\sqrt{2x+1}}{x+\sqrt{2x+1}})=0$
$\Leftrightarrow (x-2)\left [ \frac{1}{\sqrt{x+2}+2}+\frac{1}{x+\sqrt{2x+1}} \right ]=0$
$\Leftrightarrow x=2$
Vì $\frac{1}{\sqrt{x+2}+2}+\frac{1}{x+\sqrt{2x+1}}=\frac{x+2+\sqrt{x+2}+\sqrt{2x+1}}{(\sqrt{x+2}+2)(x+\sqrt{2x+1})}$ có tử >0

[tungteng532000]

 

Bài 4

$x\sqrt{2x+3}+3(\sqrt{x+5}+1)=3x+\sqrt{2x^2+3x+15}+\sqrt{2x+3}$

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