M = \(\dfrac{3^{205}+28}{3^{203}+2};\dfrac{ }{ }\) N = \(\dfrac{3^{204}+28}{3^{202}+1}\)
So sanh M và N
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so sanh M va N biet M= 3^205+28/3^203+2 N=3^204+19/3^202+1
so sánh
M= 3^205+28/3^203+2
N= 3^204+19/3^202+1
Tính 3^205 + 28 / 3^203+2
so sanh 3^205+28/3^203+2 và 3^204+19/3^202+1
\(\dfrac{2x-1}{203}\)+\(\dfrac{2x-3}{205}\)=\(\dfrac{5-2x}{207}\)-\(\dfrac{2x}{101}\)
Lời giải:
PT $\Leftrightarrow \frac{2x-1}{203}+1)+(\frac{2x-3}{205}+1)=(\frac{5-2x}{207}-1)-(\frac{2x}{101}+2)+5$
$\Leftrightarrow \frac{2x+202}{203}+\frac{2x+202}{205}=\frac{-(2x+202)}{207}-\frac{2x+202}{101}+5$
$\Leftrightarrow (2x+202)(\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})=5$
$\Leftrightarrow x=\frac{1}{2}[5: (\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})-202]$
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\(\dfrac{2x-1}{203}\)+\(\dfrac{2x-3}{205}\)=\(\dfrac{5-2x}{207}\)-\(\dfrac{2x}{101}\)
Lời giải:
PT $\Leftrightarrow \frac{2x-1}{203}+1)+(\frac{2x-3}{205}+1)=(\frac{5-2x}{207}-1)-(\frac{2x}{101}+2)+5$
$\Leftrightarrow \frac{2x+202}{203}+\frac{2x+202}{205}=\frac{-(2x+202)}{207}-\frac{2x+202}{101}+5$
$\Leftrightarrow (2x+202)(\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})=5$
$\Leftrightarrow x=\frac{1}{2}[5: (\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})-202]$
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\(\dfrac{201-x}{99}+\dfrac{203-x}{97}=\dfrac{205-x}{95}+3=0\)
`(201-x)/99+(203-x)/97+(205-x)/95+3=0`
`<=>(201-x)/99+1+(203-x)/97+1+(205-x)/95+1=0`
`<=>(300-x)/99+(300-x)/97+(300-x)/95=0`
`<=>(300-x)(1/99+1/97+1/95)=0`
`<=>300-x=0`
`<=>x=300`
Vậy `x=300`
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Vì 205-x/95+3=0
=>205-x/95=0-3=-3
=>205-x=-3*95=-285
=>x=205-(-285)=490
Vậy x=490
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\(\dfrac{201-x}{99}\)+\(\dfrac{203-x}{97}\)+\(\dfrac{205-x}{95}\)+3=0
\(\Leftrightarrow\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)
=>300-x=0
=>x=300
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<=> 300-x/99+300-x/97+300-x/95=0
<=>(300-x).(1/99+1/97+1/95)=0
<=>300-x=0
<=>x=300
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