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Formula

Calculate the value

Answer

$\dfrac{ \sqrt{ 6 } }{ 9 } - \dfrac{ \sqrt{ 24 } }{ 9 } + \dfrac{ \sqrt{ 7 } }{ 3 }$

$\dfrac { - \sqrt{ 6 } + 3 \sqrt{ 7 } } { 9 }$

Calculate the value

$\dfrac { \sqrt{ 6 } } { 9 } - \dfrac { \sqrt{ \color{#FF6800}{ 24 } } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $

$\dfrac { \sqrt{ 6 } } { 9 } - \dfrac { \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 6 } } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$\color{#FF6800}{ \dfrac { \sqrt{ 6 } } { 9 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 \sqrt{ 6 } } { 9 } } + \dfrac { \sqrt{ 7 } } { 3 }$

$ $ Since the denominator is the same as $ 9 $ , combine the fractions into one $ $

$\color{#FF6800}{ \dfrac { \sqrt{ 6 } - \left ( 2 \sqrt{ 6 } \right ) } { 9 } } + \dfrac { \sqrt{ 7 } } { 3 }$

$\dfrac { \sqrt{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 6 } } \right ) } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$ $ Get rid of unnecessary parentheses $ $

$\dfrac { \sqrt{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 6 } } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$\dfrac { \sqrt{ \color{#FF6800}{ 6 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 6 } } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$ $ Calculate between similar terms $ $

$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \sqrt{ \color{#FF6800}{ 6 } } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } \sqrt{ 6 } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$ $ Multiplying any number by 1 does not change the value $ $

$\dfrac { - \sqrt{ 6 } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$\dfrac { \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 6 } } } { 9 } + \dfrac { \sqrt{ 7 } } { 3 }$

$ $ Move the minus sign to the front of the fraction $ $

$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 6 } } { 9 } } + \dfrac { \sqrt{ 7 } } { 3 }$

$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \sqrt{ 6 } } { 9 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \sqrt{ 7 } } { 3 } }$

$ $ Find the sum of the fractions $ $

$\color{#FF6800}{ \dfrac { - \sqrt{ 6 } + 3 \sqrt{ 7 } } { 9 } }$

Solution search results

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}

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Using the \emph{removal of first derivative} method, the differential equation $ \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as $. For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$

Calculus

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Question $26$ Not answered Marked out of $1.00$ $0$ $11$ $b\right)\left(c+a$ If $a,b,c>0$ and $1\left($ \frac{ $a\right)\left(b+c\right)=1+$ $=1fraC\left($ $c\right)\left(a+b\right)1\right)$ then each fraction is equal to $b=$ Select $One$ a. $114$ $○$ b. $113$ $○$ c. $112$ $O$ d. 1 $O$

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The rationalizing factor of \sqrt{23} is $°$ $Options^{°}$ $0$ A 24 23 C \sqrt{23} D None of these

7th-9th grade

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$1f\left(u=\left(rac\left(x^{n}3+y^{n}3\right)\left(x+y\right)\right)$ then $\left($ xfrac{partial $1\right)$ {partial $x\right)+y\left(raC\left(paF|a|$ u} {partial $y\right)=2\right)$ $0$ $0$ $20$ $301$

Calculus

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