Toán Lớp 9: a) 1/(√3+√2)-√6/√3+3√(1/3) b) √(5-2√6 .)(√3+√2) Giups mik với ạ

Question

Toán Lớp 9:  a) 1/(√3+√2)-√6/√3+3√(1/3) b) √(5-2√6  .)(√3+√2) Giups mik với ạ

ngochuong2k2 · Answer

a)   1/( sqrt{3} +  sqrt{2}) - ( sqrt{6})/( sqrt{3}) + 3  sqrt{1/3}   =  1/( sqrt{3} +  sqrt{2}) -  sqrt{2} +  sqrt{3}    = (1 -  sqrt{2} ( sqrt{3} +  sqrt{2}) +  sqrt{3} (  sqrt{3} +  sqrt{2}))/( sqrt{3} +  sqrt{2})     = (1 -  sqrt{6} -  sqrt{4} +  sqrt{9} +  sqrt{6})/( sqrt{3} +  sqrt{2})     = (1 -  sqrt{4} +  sqrt{9})/( sqrt{3} +  sqrt{2})    = (1- 2 + 3)/( sqrt{3} +  sqrt{2})     = 2/( sqrt{3} +  sqrt{2})    = 2  sqrt{3} - 2  sqrt{2}   b)    sqrt{5 - 2  sqrt6}  ( sqrt{3} +  sqrt{2})    =  sqrt{ 3 - 2  sqrt6} + 2 }   ( sqrt{3} +  sqrt{2})   =   sqrt{ ( sqrt{3})^2 - 2 .  sqrt{3} .  sqrt{2} + ( sqrt{2})^2 }   ( sqrt{3} +  sqrt{2})   =   sqrt{ ( sqrt{3} -  sqrt{2})^2 }   ( sqrt{3} +  sqrt{2})   =  | sqrt{3} -  sqrt{2}|   ( sqrt{3} +  sqrt{2})   =  ( sqrt{3} -  sqrt{2})  ( sqrt{3} +  sqrt{2})   =  ( sqrt{3})^2 - ( sqrt{2})^2    = 3-2   =  1