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SovetSndri
@SovetSndri
July 2022
1
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Прошу помочь!:) представьте выражение в виде дроби (6,8,10,12)
ПОДРОБНО РЕШИТЬ!!!
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sedinalana
Verified answer
6
1)x/y+y/x-2=(x²+y²-2xy)/xy=(x-y)²/xy
2)(x-y)²/xy*xy/(4(x-y)²=1/4
8
1)b²/5-25/b=(b³-125)/5b
2)(b²+5b+25)/10b*5b/(b-5)(b²+5b+25)=1/2(b-5)
10
1)a²/(2a+1)+1=(a²+2a+1)/(2a+1)=(a+1)²/(2a+1)
2)(4a²-3)/(a=1) +4=(4a²-3+4a+4)/(a+1)=(4a²+4a+1)/(a+1)=(2a+1)²/(a+1)
3)(a+1)/(2a+1)*(2a+1)²/(a+1)=2a+1
12
1)(b²+c²)/2b +c=(b²+c²+2bc)/2b=(b+c)²/2b
2)(b²+c²)/2b -c=(b²+c²-2bc)/2b=(b-c)²/2b
3)(b+c)²/2b*2b/(b-c)²=[(b+c)/(b-c)]²
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Answers & Comments
Verified answer
61)x/y+y/x-2=(x²+y²-2xy)/xy=(x-y)²/xy
2)(x-y)²/xy*xy/(4(x-y)²=1/4
8
1)b²/5-25/b=(b³-125)/5b
2)(b²+5b+25)/10b*5b/(b-5)(b²+5b+25)=1/2(b-5)
10
1)a²/(2a+1)+1=(a²+2a+1)/(2a+1)=(a+1)²/(2a+1)
2)(4a²-3)/(a=1) +4=(4a²-3+4a+4)/(a+1)=(4a²+4a+1)/(a+1)=(2a+1)²/(a+1)
3)(a+1)/(2a+1)*(2a+1)²/(a+1)=2a+1
12
1)(b²+c²)/2b +c=(b²+c²+2bc)/2b=(b+c)²/2b
2)(b²+c²)/2b -c=(b²+c²-2bc)/2b=(b-c)²/2b
3)(b+c)²/2b*2b/(b-c)²=[(b+c)/(b-c)]²