Đáp án:

Bạn xem lại bài 2b nhé phải là `1-3+3^2-3^3+....+(-3)^x=(9^1000+1)/4`

Giải thích các bước giải:

`1,(x-1)/2020+(x-2)/2019-(x-3)/2018=(x-4)/2017`

`<=>(x-1)/2020-1+(x-2)/2019-1=(x-3)/2018-1+(x-4)/2017-`

`<=>(x-2021)/2020+(x-2021)/2019=(x-2021)/2018+(x-2021)/2017`

`<=>(x-2021)(1/2020+1/2019-1/2018-1/2017)=0`

`<=>x=2021` do `1/2020+1/2019-1/2018-1/2017<0`

Vậy `x=2021`

`2a,1/(1.3)+1/(3.5)+1/(5.7)+.....+1/((2x-1)(2x+1))=49/99`

`<=>2/(1.3)+2/(3.5)+2/(5.7)+.....+2/((2x-1)(2x+1))=98/99`

`<=>1-1/3+1/3-1/5+1/5-1/7+....+1/(2x-1)-1/(2x+1)=98/99`

`<=>1-1/(2x+1)=98/99`

`<=>(2x)/(2x+1)=98/99`

`<=>198x=196x+98`

`<=>2x=98`

`<=>x=49`

Vậy `x=49`

`b,Đặt \ A=1-3+3^2-3^3+....+(-3)^x`

`=>3A=3-3^2+3^3-3^4+.....+(-3)^{x+1}`

`=>3A+A=1+(-3)^{x+1}`

`=>4A=1+(-3)^{x+1}`

`=>A=(1+(-3)^{x+1})/4`

Mà `A=(9^1000-1)/4`

`=>1+(-3)^{x+1}=9^1000-1`

`=>(-3)^{x+1}=9^1000-2`

`=>` Không có giá trị nào của x thỏa mãn đề bài.

$\dfrac{x-1}{2020}+\dfrac{x-2}{2019}-\dfrac{x-3}{2018}-\dfrac{x-4}{2017}=0$

$\Leftrightarrow \dfrac{x-2021+2020}{2020}\,\,+\,\,\dfrac{x-2021+2019}{2019}\,\,-\,\,\dfrac{x-2021+2018}{2018}\,\,-\,\,\dfrac{x-2021+2017}{2017}=0$

$\Leftrightarrow \dfrac{x-2021}{2020}\,\,+\,\,1\,\,+\,\,\dfrac{x-2021}{2019}\,\,+\,\,1\,\,-\,\,\dfrac{x-2021}{2018}\,\,-\,\,1\,\,-\,\,\dfrac{x-2021}{2017}\,\,-\,\,1=0$

$\Leftrightarrow \dfrac{x-2021}{2020}\,\,+\,\,\dfrac{x-2021}{2019}\,\,-\,\,\dfrac{x-2021}{2018}\,\,-\,\,\dfrac{x-2021}{2017}=0$

$\Leftrightarrow \left( x-2021 \right)\left( \dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017} \right)=0$

$\Leftrightarrow x-2021=0$   ( Vì $\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{2}{2017}\ne 0$ )

$\Leftrightarrow x=2021$

Bài 2:

a)$\dfrac{1}{1.3}\,+\,\dfrac{1}{3.5}\,+\,\dfrac{1}{5.7}\,+\,...+\dfrac{1}{\left( 2x-1 \right)\left( 2x+1 \right)}=\dfrac{49}{99}$

$\Leftrightarrow \dfrac{1}{2}\left( \dfrac{2}{1.3}\,+\,\dfrac{2}{3.5}\,+\,\dfrac{2}{5.7}\,+\,\dfrac{2}{\left( 2x-1 \right)\left( 2x+1 \right)} \right)=\dfrac{49}{99}$

$\Leftrightarrow \dfrac{1}{1}\,\,\,-\,\,\dfrac{1}{3}\,\,+\,\,\dfrac{1}{3}\,\,-\,\,\dfrac{1}{5}\,\,+\,\,\dfrac{1}{5}\,\,-\,\,\dfrac{1}{7}\,+...\,+\,\,\dfrac{1}{2x-1}\,\,\,\,-\dfrac{1}{2x+1}\,\,=\dfrac{98}{99}$

$\Leftrightarrow \dfrac{1}{1}-\dfrac{1}{2x+1}=\dfrac{98}{99}$

$\Leftrightarrow \dfrac{2x}{2x+1}=\dfrac{98}{99}$

$\Leftrightarrow 99\left( 2x \right)=98\left( 2x+1 \right)$

$\Leftrightarrow 2x=98$

$\Leftrightarrow x=49$

1. b) $1-3+{{3}^{2}}-{{3}^{3}}+...-{{3}^{x}}=\dfrac{{{9}^{1000}}-1}{4}$

Đặt:

$\,\,\,S=1\,\,-\,\,3\,\,+\,\,{{3}^{2}}\,\,-\,\,{{3}^{3}}\,\,+...\,\,-\,\,{{3}^{x}}$

$3S=\,3\,\,-\,\,{{3}^{2}}\,\,+\,\,{{3}^{3}}\,\,-\,\,{{3}^{4}}\,\,+...\,\,-{{3}^{x+1}}$

$\to S+3S=\left( 1-3+{{3}^{2}}-{{3}^{3}}+...-{{3}^{x}} \right)+\left( 3-{{3}^{2}}+{{3}^{3}}-{{3}^{4}}+...-{{3}^{x+1}} \right)$

$\to 4S=1-{{3}^{x+1}}$

$\to S=\dfrac{1-{{3}^{x+1}}}{4}$

$\to \dfrac{{{9}^{1000}}-1}{4}=\dfrac{1-{{3}^{x+1}}}{4}$

$\to {{9}^{1000}}-1=1-{{3}^{x+1}}$

$\to {{3}^{x+1}}=2-{{9}^{1000}}$

${{3}^{x+1}}$ chia hết cho $3$

$2$ không chia hết cho $3$

${{9}^{1000}}$ chia hết cho $3$

Nên không có giá trị $x$

Đổi đề một chút:

$S=\dfrac{1-{{3}^{x+1}}}{4}=\dfrac{1-{{9}^{1000}}}{4}$

$\to 1-{{3}^{x+1}}=1-{{9}^{1000}}$

$\to {{3}^{x+1}}={{9}^{1000}}$

$\to {{3}^{x+1}}={{3}^{2000}}$

$\to x=1999$