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Question Number 142415 by Rankut last updated on 31/May/21

(x/(x+4))=((5⌊x⌋−7)/(7⌊x⌋−5)) x=?

xx+4=5⌊x⌋−77⌊x⌋−5x=?

Answered by mr W last updated on 31/May/21

say x=n+f ((n+f)/(n+f+4))=((5n−7)/(7n−5)) 7n^2 +7fn−5n−5f=5n^2 +5(f+4)n−7n−7(f+4) f=((9n−n^2 −14)/(n+1))≥0 n^2 −9n+14≤0 (n−2)(n−7)≤0 ⇒2≤n≤7 f=((9n−n^2 −14)/(n+1))<1 n^2 −8n+15>0 (n−3)(n−5)>0 ⇒n<3 or n>5 ⇒n=2, 6, 7 f=((9×2−2^2 −14)/(2+1))=0 ⇒x=2 f=((9×6−6^2 −14)/(6+1))=(4/7) ⇒x=6(4/7) f=((9×7−7^2 −14)/(7+1))=0 ⇒x=7

sayx=n+fn+fn+f+4=5n−77n−57n2+7fn−5n−5f=5n2+5(f+4)n−7n−7(f+4)f=9n−n2−14n+1⩾0n2−9n+14⩽0(n−2)(n−7)⩽0⇒2⩽n⩽7f=9n−n2−14n+1<1n2−8n+15>0(n−3)(n−5)>0⇒n<3orn>5⇒n=2,6,7f=9×2−22−142+1=0⇒x=2f=9×6−62−146+1=47⇒x=647f=9×7−72−147+1=0⇒x=7

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