Question

Anyone know how to do this question? ​Pls help me all people

Answer 1

Try this solution, pls, modify it according to Your requirements:

$\sqrt{12+6\sqrt{3}} -\sqrt{3-2\sqrt{2}} =a+\sqrt{b} -\sqrt{c};$

$\sqrt{(3+\sqrt{3})^2} -\sqrt{(\sqrt{2} -1)^2}=a+\sqrt{b} -\sqrt{c};$

$3+\sqrt{3}-\sqrt{2}+1=a+\sqrt{b} -\sqrt{c};$

$4+\sqrt{3} -\sqrt{2} =a+\sqrt{b} -\sqrt{c};$

a=4; b=3; c=2.

additional:

$\sqrt{12+6\sqrt{3}}=\sqrt{3^2+(\sqrt{3})^2+2*3*\sqrt{3}}=\sqrt{(3+\sqrt{3} )^2};$

$\sqrt{3-2\sqrt{2}} =\sqrt{(\sqrt{2})^2-2*1*\sqrt{2}+1^2}=\sqrt{(\sqrt{2} -1)^2}.$

note, (√2-1) is the correct result instead of (1-√2). Short explanation: the result (√2-1) is positive value under the 'square root' sign, the negative values under the 'square root' sign are like restricted operation, finally only (√2-1) result is legal.