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Algebra 1A help please thank you!!

Algebra 1A help please thank you!!-example-1

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Answer:


\large\text{$√(50)=\boxed{5√(2)}$}


\large\text{$√(50)=\boxed{5\cdot 2^{(1)/(2)}}$}


\large\text{$\sqrt[10]{2^5}=\boxed{2^{(1)/(2)}}$}

Explanation:

Given expression:


\large\text{$\sqrt[10]{2^5}\cdot 5 \cdot 2^{-(1)/(2)}-\frac{√(50)}{2^{(1)/(2)}}$}

To simplify the given expression, begin by simplifying and rewriting the radicals.

To simplify the radical √(50) using perfect square factors, we can rewrite 50 as the product of 5² · 2:


\large\text{$√(50)=√(5^2 \cdot 2)$}


\textsf{Apply the radical rule:} \quad √(ab)=\sqrt{\vphantom{b}a}√(b)


\large\text{$√(50)=√(5^2) √( 2)$}


\textsf{Apply the radical rule:} \quad √(a^2)=a, \quad a \geq 0


\large\text{$√(50)=5√(2)$}

The square root can be represented as a rational exponent of 1/2. Therefore:


\large\text{$√(50)=5\cdot 2^{(1)/(2)}$}

To rewrite
\sqrt[10]{2^5}, we can apply the exponent rule
\sqrt[n]{a^m}=a^{(m)/(n)}:


\large\text{$\sqrt[10]{2^5}=2^{(5)/(10)}=2^{(1)/(2)}$}

Therefore, the expression simplified by rewriting the radicals is:


\large\text{$2^{(1)/(2)}\cdot 5 \cdot 2^{-(1)/(2)}-\frac{5\cdot 2^{(1)/(2)}}{2^{(1)/(2)}}$}

To simplify further, cancel the common factor
2^{(1)/(2)} of the rational:


\large\text{$2^{(1)/(2)}\cdot 5 \cdot 2^{-(1)/(2)}-\frac{5\cdot \diagup\!\!\!\!\! 2^{(1)/(2)}}{\diagup\!\!\!\!\!2^{(1)/(2)}}$}


\large\text{$2^{(1)/(2)}\cdot 5 \cdot 2^{-(1)/(2)}-5$}


\textsf{Apply the exponent rule:} \quad a^b \cdot a^c=a^(b+c)


\large\text{$2^{(1)/(2)}\cdot 2^{-(1)/(2)}\cdot 5 -5$}


\large\text{$2^{(1)/(2)-(1)/(2)}\cdot 5 -5$}


\large\text{$2^(0)\cdot 5 -5$}

As any number to the power of zero is always 1:


\large\text{$1\cdot 5 -5$}


\large\text{$5 -5$}


\large\text{$0$}

Therefore, the expression is equal to zero.

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