Question

√6+2√5= 1+√x
answer for this....​

Answer 1

Step-by-step explanation:

$\sqrt{6} + 2 \sqrt{5} = 1 + \sqrt{x} \\ \sqrt{6} + 2 \sqrt{5} - 1 = \sqrt{x} \\ squaring \: on \: both \: sides \\ {( \sqrt{6 } + 2 \sqrt{5} - 1) }^{2} = {( \sqrt{x}) }^{2} \\ using \: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ { (\sqrt{6}) }^{2} + {(2 \sqrt{5})}^{2} + {( - 1)}^{2} + 2 \times \sqrt{6} \times 2 \sqrt{5} + 2 \times 2 \sqrt{5} \times ( - 1) + 2 \times ( - 1) \times \sqrt{6} = x \\ 6 + (4 \times 5) + 1 + 4 \sqrt{30} + ( - 4 \sqrt{5} ) + ( - 2 \sqrt{6} ) = x \\ 6 + 20 + 1 + 4 \sqrt{30} - 4 \sqrt{5} - 2 \sqrt{6} = x \\ 27 + 4 \sqrt{30} - 4 \sqrt{5} - 2 \sqrt{6} = x$

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