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4 - 1 = 3 4 - 2 = 2 3 + 1 = 4 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 4 - 3 = 1 3 - 1 = 2
2 - 1 = 1 4 - 3 = 1 4 - 1 = 3 3 - 2 = 1
Lời giải chi tiết:
| 4 – 1 > 2 | 4 – 3 < 4 – 2 |
| 4 – 2 = 2 | 4 – 1 < 3 + 1 |
| 3 – 1 = 2 | 3 – 1 > 3 – 2 |
Lời giải chi tiết:
| 4 – 1 > 2 | 4 – 3 < 4 – 2 |
| 4 – 2 = 2 | 4 – 1 < 3 + 1 |
| 3 – 1 = 2 | 3 – 1 > 3 – 2 |
a: =>x-2/5=3/4:1/3=3/4*3=9/4
=>x=9/4+2/5=45/20+8/20=53/20
b: =>x-2/3=7/3:4/5=7/3*5/4=35/12
=>x=35/12+2/3=43/12
c: 1/3(x-2/5)=4/5
=>x-2/5=4/5*3=12/5
=>x=12/5+2/5=14/5
d: =>2/3x-1/3-1/4x+1/10=7/3
=>5/12x-7/30=7/3
=>5/12x=7/3+7/30=77/30
=>x=77/30:5/12=154/25
e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)
=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)
=>x=19/7:23/28=76/23
f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5
=>13/12x=1/5+3/2+4/3+5/4=257/60
=>x=257/65
i: =>x^2-2/5x-x^2-2x+11/4=4/3
=>-12/5x=4/3-11/4=-17/12
=>x=17/12:12/5=85/144
a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(A=1-\frac{1}{2^{50}}<1\)
Vậy \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}<1\)
b)\(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3B-B=2B=1-\frac{1}{3^{100}}\)
\(B=\frac{1-\frac{1}{3^{100}}}{2}\)
Vì \(1-\frac{1}{3^{100}}<1\)nên\(\frac{1-\frac{1}{3^{100}}}{2}<\frac{1}{2}\)
Vậy \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}<\frac{1}{2}\)
c) \(C=\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\)
\(4C=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{998}}+\frac{1}{4^{999}}\)
\(4C-C=3C=1-\frac{1}{4^{1000}}\)
\(C=\frac{1-\frac{1}{4^{1000}}}{3}\)
Vì \(1-\frac{1}{4^{1000}}<1\)nên\(\frac{1-\frac{1}{4^{1000}}}{3}<\frac{1}{3}\)
Vậy \(C=\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}<\frac{1}{3}\)