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14 tháng 10 2023

\(x^{99}-x^{97}=0\Leftrightarrow x^{97}\cdot\left(x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^{97}=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

25 tháng 9 2016

x=100

Ta sẽ có: 1-1+1+1-1+1-1+1=0

25 tháng 9 2016

\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)

\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)

\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)

Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)

\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

25 tháng 9 2016

x thuoc R

15 tháng 7 2019

a) x - 3/97 + x - 2/98 = x - 1/99 + x/100

<=> x + 1/99 + 1 + x + 2/98 + 1 + x + 3/97 + 1 + (x + 4/96 + 1 + x + 5/95 + 1 + x + 10/90 + 1) = 0

<=> x + 100/99 + x + 100/98 + x + 100/97 + (x + 100/96 + x + 100/95 + x + 100/90) = 0

<=> (x + 100)(1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90) = 0

Mà 1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90 khác 0

=> x + 100 = 0

=> x = -100

c) (1/1.2 + 1/2.3 + ... + 1/99.100) - 2x = 1/2

<=> (1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100) - 2x = 1/2

<=> (1 - 1/100) - 2x = 1/2

<=> 99/100 - 2x = 1/2

<=> -2x = 1/2 - 99/100

<=> -2x = -49/100

<=> x = 49/200

=> x = 49/200

15 tháng 7 2019

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

30 tháng 12 2016

(x+5)/100+(x+5)/99=(x+5)/98+(x+5)/97

(x+5)/100+(x+5)/99-(x+5)/98-(x+5)/97=0

(x+5)(1/100+1/99-1/98-1/97)=0

suy ra x+5=0

x=-5

10 tháng 7 2015

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=0\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)

\(\Leftrightarrow x+100=0\text{ (do }\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\ne0\text{)}\)

\(\Leftrightarrow x=-100\)

10 tháng 7 2015

bn tự chép đề lại nha

từ đề bài suy ra  \(1+\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+3=1+1+1+0=3\)

suy ra \(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+3=3\)

suy ra \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=3-3=0\)

suy ra \(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

mà 1/99 +1/98+1/97 lớn hơn 0

từ 2 điều trên suy ra x+100=0 suy ra x=-100

28 tháng 11 2016

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)