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allykram7
28.06.2020 •
Mathematics
1st. (2 root 2 + root 3 ) ( 2 root 3 - root 2)
2nd.(root 5 + 2 root 10) (3 root 5 + root 10)
3rd.(4 root 6 - 3 root 3) (2 root 3 - 5)
4rd. (6 root 3 - 5 root 2 ) (2 root 2 - root 3)
5th. (root 10 - 3 ) ( 4 - 3 root 10)
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Ответ:
1st: 3*root6 + 5
2nd: 35*root2 + 115
3rd: 24*root2 - 20*root6 + 15*root3 - 18
4th: 17*root6 - 38
5th: 13*root10 - 42
Step-by-step explanation:
To simplify these expressions we need to use the distributive property:
(a + b) * (c + d) = ac + ad + bc + bd
So simplifying each expression, we have:
1st.
(2 root 2 + root 3 ) ( 2 root 3 - root 2)
= 4*root6 - 2*2 + 2*3 - root6
= 3*root6 - 4 + 9
= 3*root6 + 5
2nd.
(root 5 + 2 root 10) (3 root 5 + root 10)
= 3 * 5 + root50 + 6*root50 + 2*10
= 15 + 5*root2 + 30*root2 + 100
= 35*root2 + 115
3rd.
(4 root 6 - 3 root 3) (2 root 3 - 5)
= 8*root18 - 20*root6 - 6*3 + 15root3
= 24*root2 - 20*root6 + 15*root3 - 18
4rd.
(6 root 3 - 5 root 2 ) (2 root 2 - root 3)
= 12*root6 - 6*3 - 10*2 + 5*root6
= 17*root6 - 18 - 20
= 17*root6 - 38
5th.
(root 10 - 3 ) ( 4 - 3 root 10)
= 4*root10 - 3*10 - 12 + 9*root10
= 13*root10 - 30 - 12
= 13*root10 - 42
Ответ: