Question

Am nevoie de a,b,c de la ex 7. Va rog !

Answer 1

Răspuns:

Explicație pas cu pas:

a) (5√2/2√5)+(3/√10)-(2√5/5√2)-(3√5/√8)=

(5√2/2√5)+(3/√10)-(2√5/5√2)-(3√5/2√2)=

(5√2·√5/2·5)+(3·√10/10)-(2√5·√2/5·2)-(3√5·√2/2·2)=

(5√10/10)+(3√10/10)-(2√10/10)-(3√10/4)=(5√10+3√10-2√10)/10-(3√10/4)=

(6√10/10)-(3√10/4)=(3√10/5)-(3√10/4)=(4·3√10-5·3√10)/20=

(12√10-15√10)/20=-3√10/20

b) (2/√6)+(3√2/2√3)-[(2√3/3√2)+(8√2/√27)]=      

(2/√6)+(3√2/2√3)-[(2√3/3√2)+(8√2/3√3)]=

(2√6/6)+(3√2·√3/2·3)-[(2√3·√2/3·2)+(8√2·√3/3·3)]=  

(2√6/6)+(3√6/6)-[(2√6/6)+(8√6/9)]=(5√6/6)-[(3·2√6+2·8√6)/18]=

(5√6/6)-(22√6/18)=(3·5√6-22√6)/18=-7√6/18          

c) (2√2/3√3)+5√2/2√3-[(7√2/6√3)+(4√2/√3)=

(2√2·√3/3·3)+(5√2·√3/2·3)-[(7√2·√3/6·3)+(4√2·√3/3)]=

(2√6/9)+(5√6/6)-[(7√6/18)+(4√6/3)]=(2·2√6+3·5√6)/18-(7√6+6·4√6)/18=

(19√6/18)-(31√6/18)=-12√6/18=-2√6/3