Question

DAU COROANA!!!

calculati:

$2x \sqrt{ ( \sqrt{ 5 } + \sqrt{ 3 } ^ { 2 } } -(2 \sqrt{ 3 } - \sqrt{ 5 } )x=x \sqrt{ 20 } + \sqrt{ 125 }$
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Answer 1

Răspuns:

Explicație pas cu pas:

$2x \sqrt{( \sqrt{5} + \sqrt{3} ) {}^{2} } - (2 \sqrt{3} - \sqrt{5} )x = x\sqrt{20} + \sqrt{125}$

$2x( \sqrt{5} + \sqrt{3} ) - (2 \sqrt{3} x - \sqrt{5} x) = x \times 2 \sqrt{5} + 5 \sqrt{5}$

$2 \sqrt{5} x + 2 \sqrt{3} x - (2 \sqrt{3} x - \sqrt{5} x) = 2 \sqrt{5} x + 5 \sqrt{5}$

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

$2 \sqrt{5} x + \sqrt{5} x = 2 \sqrt{5} x + 5 \sqrt{5} \\ 2 \sqrt{5} x - 2 \sqrt{5} x + \sqrt{5} x = 5 \sqrt{5}$

$\sqrt{5} x = 5 \sqrt{5}$

$x = \frac{5 \sqrt{5} }{ \sqrt{5} } = \boxed{5}$

Answer 2

$\it 2x\sqrt{(\sqrt5-\sqrt3)^2}-(2\sqrt3-\sqrt5)x=x\sqrt{20}+\sqrt{125} \Rightarrow 2x|\underbrace{\it\sqrt5+\sqrt3}_{ > 0}|-\\ \\ -2x\sqrt3+x\sqrt5=x\sqrt{4\cdot5}+\sqrt{25\cdot5} \Rightarrow \underline{2x\sqrt5}+\underline{\underline{2x\sqrt3}}-\underline{\underline{2x\sqrt3}}+\\ \\ +x\sqrt5-\underline{2x\sqrt5}=5\sqrt5 \Rightarrow x\sqrt5=5\sqrt5 \Rightarrow x=5$