Question

calculati: 15 supra 8 radical din 2-6 radical din 3+ 20 supra 7 radical din 2-4 radical din 3- 4 supra 5 radical din 3-6 radical din 2. va rog sa ma ajutati !! ESTE IMPORTANT...eu am rezolvat operatia, dar nu stiu daca este rezultatul corect....defapt, stiu ca nu este corect...si as avea nevoie de ajutor...nu ignorati, MULTUMESC !

Answer 1

15/(8√2 -6√3) +20/(7√2 -4√3 )-4/(5√3-6√2)=
=15(8√2 +6√3)/(8√2  -6√3)(8√2 +6√3) +20(7√2 +4√3 )/(7√2 -4√3 )(7√2 +4√3 ) --4(5√3+6√2)/(5√3-6√2)(5√3+6√2)=
=(120√2 +90√3)/(128-108) +(140√2 +80√3 )/(98 -48)-(20√3+24√2)/(75-72)=
=(120√2 +90√3)/20 +(140√2 +80√3 )/50-(20√3+24√2)/3=
=(12√2 +9√3)/2 +(14√2 +8√3 )/5-(20√3-24√2)/3=
=(180√2 +135√3 +84√2 +48√3 -200√3-240√2)/30=
=(24√2-17√3)/30

Answer 2

$\displaystyle \frac{15}{8 \sqrt{2} -6 \sqrt{3} } + \frac{20}{7 \sqrt{2} -4 \sqrt{3} } - \frac{4}{5 \sqrt{3} -6 \sqrt{2} } = \\ \\ = \frac{15(8 \sqrt{2} +6 \sqrt{3} )}{128-108} + \frac{20(7 \sqrt{2} +4 \sqrt{3} )}{98-48} - \frac{4(5 \sqrt{3} +6 \sqrt{2} )}{75-72} = \\ = \frac{15(8 \sqrt{2} +6 \sqrt{3} )}{20} + \frac{20(7 \sqrt{2} +4 \sqrt{3} )}{50} - \frac{4(5 \sqrt{3} +6 \sqrt{2} )}{3} =$

$\displaystyle = \frac{3(8 \sqrt{2}+6 \sqrt{3}) }{4} + \frac{2(7 \sqrt{2} +4 \sqrt{3} )}{5} - \frac{4(5 \sqrt{3} +6 \sqrt{2}) }{3} = \\ \\ = \frac{24 \sqrt{2} +18 \sqrt{3} }{4} + \frac{14 \sqrt{2} +8 \sqrt{3} }{5} - \frac{20 \sqrt{3} +24 \sqrt{2} }{3} = \\ \\ = \frac{15(24 \sqrt{2} +18 \sqrt{3} )+12(14 \sqrt{2} +8 \sqrt{3} )-20(20 \sqrt{3} +24 \sqrt{2} )}{60} =$

$\displaystyle = \frac{360 \sqrt{2}+270 \sqrt{3} +168 \sqrt{2} + 96 \sqrt{3} -400 \sqrt{3} -480 \sqrt{2} }{60} = \\ \\ = \frac{48 \sqrt{2}- 34 \sqrt{3} }{60} = \frac{2(24 \sqrt{2} -17 \sqrt{3} )}{60} = \frac{24 \sqrt{2} -17 \sqrt{3} }{30}$