Question

calculați x+y și x-y, pentru numerele : x=√8 - √50+√27 și y= √48 + √98-√200;
b) x=√49 - 3√20 + √28 și y= √180 -√112+√64.
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Answer 1

Răspuns:

a) x + y = - 6√2 + 7√3 ; x - y = - √3

b) x + y = 15 - 2√7 ; x - y = - 1 - 12√5 + 6√7

Explicație pas cu pas:

x + y = ?

x - y = ?

a) x = √8 - √50 + √27

x = 2√2 - 5√2 + 3√3

x = - 3√2 + 3√3

y = √48 + √98 - √200

y = 4√3 + 7√2 - 10√2

y = 4√3 - 3√2

x + y = - 3√2 + 3√3 + 4√3 - 3√2

x + y = - 6√2 + 7√3

x - y = - 3√2 + 3√3 - ( 4√3 - 3√2 )

x - y = - 3√2 + 3√3 - 4√3 + 3√2

x - y = - √3

b) x = √49 - 3√20 + √28

x = 7 - 3 × 2√5 + 2√7

x = 7 - 6√5 + 2√7

y = √180 - √112 + √64

y = 6√5 - 4√7 + 8

x + y = 7 - 6√5 + 2√7 + 6√5 - 4√7 + 8

x + y = 15 - 2√7

x - y =  7 - 6√5 + 2√7 - ( 6√5 - 4√7 + 8 )

x - y =  7 - 6√5 + 2√7 - 6√5 + 4√7 - 8

x - y = - 1 - 12√5 + 6√7.

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Answer 2

a)

$x=\sqrt{8} -\sqrt{50} +\sqrt{27} \\x=2\sqrt{2} -5\sqrt{2} +3\sqrt{3} \\x=3\sqrt{3} -3\sqrt{2} \\\\y=\sqrt{48}+\sqrt{98}-\sqrt{200}\\y= 4\sqrt{3} + 7\sqrt{2} -10\sqrt{2} \\y=4\sqrt{3} -3\sqrt{2} \\\\x+y=3\sqrt{3} -3\sqrt{2} +4\sqrt{3} -3\sqrt{2}\\x+y=7\sqrt{3} -6\sqrt{2}\\\\x-y=3\sqrt{3} -3\sqrt{2} -(4\sqrt{3}-3\sqrt{2} )\\x-y=3\sqrt{3} -3\sqrt{2} -4\sqrt{3} +3\sqrt{2} \\x-y=-\sqrt{3}$

b)

$x=\sqrt{49}-3\sqrt{20}+\sqrt{28} \\x=7-6\sqrt{5}+2\sqrt{7}\\\\y=\sqrt{180}-\sqrt{112}+\sqrt{64} \\y=6\sqrt{5} -4\sqrt{7 }+8\\\\x+y=7-6\sqrt{5}+2\sqrt{7}+6\sqrt{5} -4\sqrt{7 }+8\\x+y=15-2\sqrt{7} \\\\x-y=7-6\sqrt{5}+2\sqrt{7}-(6\sqrt{5} -4\sqrt{7 }+8)\\x-y=7-6\sqrt{5}+2\sqrt{7}-6\sqrt{5} +4\sqrt{7} -8\\x-y=-1-12\sqrt{5} +6\sqrt{7} \\x-y=6\sqrt{7} -12\sqrt{5} -1$

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