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Maksimus2004
@Maksimus2004
July 2022
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ПОЖАЛУЙСТА ПОМОГИТЕ решить № 1, 2 И 3
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daniilgrighorip6inck
1)
a)
4a/a²-1 + a-1/a+1 = 4a/(a-1)·(a+1) + a-1/a+1 = 4a+(a-1)²/(a-1)·(a+1) = 4a+a²-2a+1/(a-1)·(a+1) = 2a+a²+1/(a-1)·(a+1) = a²+2a+1/(a-1)·(a+1) = (a+1)²/(a-1)·(a+1) = a+1/a-1
б)2b-5/b²-5b + 1/5-b = 2b-5/b·(b-5) + 1/-(b-5) = 2b-5/b·(b-5) - 1/b-5 = 2b-5-b/b·(b-5) = b-5/b·(b-5) = 1/b
в)12x/x²-9 + x-3/x+3 = 12x/(x-3)·(x+3) + x-3/x+3 = 12x+(x-3)²/(x-3)·(x+3) = 12x+x²-6x+9/(x-3)·(x+3) = 6x+x²+9/(x-3)·(x+3) = x²+6x+9/(x-3)·(x+3) = (x+3)²/(x-3)·(x+3) = x+3/x-3
г)m+2/3m²-3m - 1/m-1 = m+2/3m·(m-1) - 1/m-1 = m+2-3m/3m·(m-1) = -2m+2/3m·(m-1) = -2(m-1)/3m·(m-1) = -2/3m = - 2/3m
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Answers & Comments
a)4a/a²-1 + a-1/a+1 = 4a/(a-1)·(a+1) + a-1/a+1 = 4a+(a-1)²/(a-1)·(a+1) = 4a+a²-2a+1/(a-1)·(a+1) = 2a+a²+1/(a-1)·(a+1) = a²+2a+1/(a-1)·(a+1) = (a+1)²/(a-1)·(a+1) = a+1/a-1
б)2b-5/b²-5b + 1/5-b = 2b-5/b·(b-5) + 1/-(b-5) = 2b-5/b·(b-5) - 1/b-5 = 2b-5-b/b·(b-5) = b-5/b·(b-5) = 1/b
в)12x/x²-9 + x-3/x+3 = 12x/(x-3)·(x+3) + x-3/x+3 = 12x+(x-3)²/(x-3)·(x+3) = 12x+x²-6x+9/(x-3)·(x+3) = 6x+x²+9/(x-3)·(x+3) = x²+6x+9/(x-3)·(x+3) = (x+3)²/(x-3)·(x+3) = x+3/x-3
г)m+2/3m²-3m - 1/m-1 = m+2/3m·(m-1) - 1/m-1 = m+2-3m/3m·(m-1) = -2m+2/3m·(m-1) = -2(m-1)/3m·(m-1) = -2/3m = - 2/3m