Tính giá trị của biểu thức :

B

baochauhn1999

Ta có:
$x^3=\sqrt{2}-1-\frac{1}{\sqrt{2}-1}-3x$
Cho nên:
$P=\sqrt{2}-1-\frac{1}{\sqrt{2}-1}-3x+3x+2=\sqrt{2}+1+\frac{1}{\sqrt{2}-1}$
 
E

eye_smile

$a=\sqrt[3]{\sqrt{2}-1}$; $b=\dfrac{1}{\sqrt[3]{\sqrt{2}-1}}$
\Rightarrow $x=a-b$; $ab=1$
Có: ${x^3}+3x+2={(a-b)^3}+3x+2={a^3}-{b^3}-3ab(a-b)+3x+2=\sqrt{2}-1-\dfrac{1}{\sqrt{2}-1}-3x+3x+2=\sqrt{2}-1-\sqrt{2}-1-3x+3x+2=0$
 
D

duchieu300699

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