

s=1/2 + 1/14 + 1/35 + 1/65 + ... + 1/4850. chứng minh rằng S<2/3
Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
Đáp án:
`S=1/2 + 1/14 + 1/35 + 1/65 + ... + 1/4850`
`S=2/4 + 2/28 + 2/70 + 2/130 + ... + 2/9700`
`S=2/(1*4) + 2/(4*7) + 2/(7*10) + 2/(10*13) + ... + 2/(97*100)`
`3/2 *S=3/2 * (2/(1*4) + 2/(4*7) + 2/(7*10) + 2/(10*13) + ... + 2/(97*100))`
`3/2 *S= 3/(1*4) + 3/(4*7) + 3/(7*10) + 3/(10*13) + ... + 3/(97*100)`
`3/2 *S= 1/1-1/4 + 1/4-1/7 + 1/7-1/10 + 1/10*-1/13 + ... + 1/97-1/100`
`3/2 *S= 1/1-1/100`
`3/2 *S= 99/100`
`S= 99/100:3/2`
`S= 99/100*2/3`
`S= 33/50`
`S= 99/150`
`S<100/150`
`S<2/3`
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`**` `**` `**` ` Câu ` `1` `` `A=(2/3-0,75):sqrt(1/16)+[(-2)/15:((-2)/5)^2]sqrt(64/25)+6sqrt(25/144)` `` `A=(2/3-3/4):1/4+[(-2)/15:4/25]*8/5+6*5/12` `` `A=(-1)/12*4+(-5)/15*25/4*8/5+5/2` `` `A=(-1)/3-10/3+5/2` `` `A=(-11)/3 + 5/2` `` `A=(-22)/6+15/6` `` `A=-7/6` Rút gọn`**` `**` `**` ` Câu ` `1` `` `A=(2/3-0,75):sqrt(1/16)+[(-2)/15:((-2)/5)^2]sqrt(64/25)+6sqrt(25/144)` `` `A=(2/3-3/4):1/4+[(-2)/15:4/25]*8/5+6*5/12` `` `A=(-1)/12*4+(-5)/15*25/4*8/5+5/2` `` `A=(-1)/3-10/3+5/2` `` `A=(-11)/3 + 5/2` `` `A=(-22)/6+15/6` `... xem thêm
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`**` `**` `A=1+3^2+3^4+....+3^2008` `` `3^2A=3^2+3^4+3^6+...+3^2010` `` `9A-A=(3^2+3^4+3^6+...+3^2010)` `` `-(1+3^2+3^4+....+3^2008)` `` `8A=3^2010-1` `` `=>B=8A-3^2010=3^2010-1-3^2010=-1`
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`B=32008−1−32010B = 3^{2008} - 1 - 3^{2010}` `B=32008−1−32008+2B = 3^{2008} - 1 - 3^{2008+2}` `B=32008−1−32008⋅32B = 3^{2008} - 1 - 3^{2008} \cdot 3^2` `B=32008−1−9⋅32008B = 3^{2008} - 1 - 9 \cdot 3^{2008}` `B=32008−9⋅32008−1B = 3^{2008} - 9 \cdot 3^{2008} - 1` `B=32008(1−9)−1B = 3^{2008}(1 - 9) - 1` `B=32008(−8)−1B = 3^{2008}(-8) - 1` `B=−8⋅32008−1B = -8 \cdot 3^{2008} - 1` Vậy giá trị của biểu thức B là: Rút gọn`B=32008−1−32010B = 3^{2008} - 1 - 3^{2010}` `B=32008−1−32008+2B = 3^{2008} - 1 - 3^{2008+2}` `B=32008−1−32008⋅32B = 3^{2008} - 1 - 3^{2008} \cdot 3^2` `B=32008−1−9⋅32008B = 3^{2008} - 1 - 9 \cdot 3^{2008}` `B=32008−9⋅32008−1B = 3^{2008} - 9 \cdot 3^{2... xem thêm
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`(-2)/x=y/5` `xy=-10` `xy=-10=-5.2=-2.5` ( vì `x<0 < y` )` `=>(x;y) in {(-5;2);(-2;5)}`
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`(-2)/x=y/5` `xy=-10` `xy=-10=-5.2=-2.5` ( vì `x<0 < y` ) `=>(x;y) in {(-5;2);(-2;5)}`
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** `**` `27/81 ; 25/100 ; 35/84` `` `27/81=1/3` `` `25/100=1/4` `` `35/84=5/12` `` `=>` `MCS : 12` `` `1/3=(1xx4)/4xx3)=4/12` `` `1/4=(1xx3)/(4xx3)=3/12` `` `5/12`
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`27/81 ; 25/100 ; 35/84` `` `27/81=1/3` `` `25/100=1/4` `` `35/84=5/12` `` `=>` `MCS : 12` `` `1/3=(1xx4)/(4xx3)=4/12` `` `1/4=(1xx3)/(4xx3)=3/12` `` `5/12`
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`` `` `a)5x-(2-3x)=4(x+3)` `` `5x-2+3x=4x+12` `` `5x+3x-4x=12+2` `` `4x=14` `` `x=7/2` `` `c)5x-(2-4x)=6+3(x-1)` `` `5x-2+4x=6+3x-3` `` `9x-2=3+3x` `` `9x-3x=3+2` `` `6x=5` `` `x=5/6`