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Formula
Calculate the value
Answer
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$\dfrac{ 2 }{ 1- \sqrt{ 2 } }$
$- 2 - 2 \sqrt{ 2 }$
Calculate the value
$\dfrac { 2 } { 1 - \sqrt{ 2 } }$
$ $ Find the conjugate irrational number of denominator $ $
$\color{#FF6800}{ \dfrac { 2 } { 1 - \sqrt{ 2 } } } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 + \sqrt{ 2 } } { 1 + \sqrt{ 2 } } }$
$\dfrac { 2 } { 1 - \sqrt{ 2 } } \times \dfrac { 1 + \sqrt{ 2 } } { 1 + \sqrt{ 2 } }$
$ $ The denominator is multiplied by denominator, and the numerator is multiplied by numerator $ $
$\color{#FF6800}{ \dfrac { 2 \left ( 1 + \sqrt{ 2 } \right ) } { \left ( 1 - \sqrt{ 2 } \right ) \left ( 1 + \sqrt{ 2 } \right ) } }$
$\dfrac { \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 2 } } \right ) } { \left ( 1 - \sqrt{ 2 } \right ) \left ( 1 + \sqrt{ 2 } \right ) }$
$ $ Multiply each term in parentheses by $ 2$
$\dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } } { \left ( 1 - \sqrt{ 2 } \right ) \left ( 1 + \sqrt{ 2 } \right ) }$
$\dfrac { 2 + 2 \sqrt{ 2 } } { \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \right ) \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ + } \sqrt{ \color{#FF6800}{ 2 } } \right ) }$
$ $ Expand the expression using $ \left(a - b\right)\left(a + b\right) = a^{2} - b^{2}$
$\dfrac { 2 + 2 \sqrt{ 2 } } { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \sqrt{ \color{#FF6800}{ 2 } } \right ) ^ { \color{#FF6800}{ 2 } } }$
$\dfrac { 2 + 2 \sqrt{ 2 } } { \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } } - \left ( \sqrt{ 2 } \right ) ^ { 2 } }$
$ $ Calculate power $ $
$\dfrac { 2 + 2 \sqrt{ 2 } } { \color{#FF6800}{ 1 } - \left ( \sqrt{ 2 } \right ) ^ { 2 } }$
$\dfrac { 2 + 2 \sqrt{ 2 } } { 1 - \left ( \sqrt{ \color{#FF6800}{ 2 } } \right ) ^ { \color{#FF6800}{ 2 } } }$
$ $ Calculate power $ $
$\dfrac { 2 + 2 \sqrt{ 2 } } { 1 - \color{#FF6800}{ 2 } }$
$\dfrac { 2 + 2 \sqrt{ 2 } } { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } }$
$ $ Subtract $ 2 $ from $ 1$
$\dfrac { 2 + 2 \sqrt{ 2 } } { \color{#FF6800}{ - } \color{#FF6800}{ 1 } }$
$\dfrac { 2 + 2 \sqrt{ 2 } } { \color{#FF6800}{ - } 1 }$
$ $ If the denominator is 1, the denominator can be removed $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } \right )$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } \right )$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } }$
Solution search results
search-thumbnail-Which of the following rational numbers are 
equivalent? 
$0Ptionsy$ 
A \frac{5}{6}, \frac{30}{36} 
B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ 
C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ 
D \frac{1}{2},\frac{3}{8}
7th-9th grade
Other
search-thumbnail-The rationalizing factor of \sqrt{23} is 
$°$ $Options^{°}$ $0$ 
A 24 
23 
C \sqrt{23} 
D None of these
7th-9th grade
Other
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