Question

PT 1÷2+√3 + 2÷√5 +√3 + 1÷2-√5 =0​

Answer 1

Step-by-step explanation:

1/2+√3)+(2/√5-√3)+(1/2-√5)

=1(2-√3)/(2-√3)(2+√3)+2(√5+√3)/(√5+√3)(√5-√3)+1(2+√5)/(2+√5)(2-√5)

=(2-√3)/{2²-(√3)²}+2(√5+√3)/{(√5)²-(√3)²}+(2+√5)/{2²-(√5)²}

=(2-√3)/(4-3)+2(√5+√3)/(5-3)+(2+√5)/(4-5)

=2-√3+{2(√5+√3)/2}-(2+√5)

=2-√3+√5+√3-2-√5

=0 (Proved)

Example ↑

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$\huge \pink{✿} \red {C} \green {u} \blue {t} \orange {e} \pink {/} \red {Q} \blue {u} \pink {e} \red {e} \green {n} \pink {♡}$

Answer 2

$\huge\boxed{\fcolorbox{red}{ink}{SOLUTION:}}$

refer the attachment I solve this in my notebook

$\huge\boxed{\dag\sf\red{Thanks}\dag}$