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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 \(x=100\)
NM
1
LL
28 tháng 11 2016
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)
\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
TN
1
22 tháng 7 2016
Đặt \(.K=\frac{x+99}{-1}=\frac{y-98}{2}=\frac{z+97}{-3}\)
\(\Rightarrow\frac{x+97}{K}=-1\)
\(\Rightarrow\frac{y-98}{K}=2\)
\(\Rightarrow\frac{z+97}{K}=-3\)
\(\Rightarrow\frac{x+99}{K}+\frac{y-98}{K}+\frac{z+97}{K}=\left(-1\right)+2+\left(-3\right)\)
\(\Rightarrow\frac{\left(x+99\right)+\left(y-98\right)+\left(z+97\right)}{K}=-2\)
Đến đây thì ... mình quên mất tiêu rồi bạn tự nghĩ tiếp nha :)
P
1
12 tháng 11 2015
\(\frac{x+99}{-1}=\frac{y-98}{2}=\frac{z+97}{-3}=\frac{x+99-\left(y-98\right)+\left(z+97\right)}{-1-2+\left(-3\right)}=\frac{\left(x-y+z\right)+294}{-6}=\frac{50+294}{-6}=-\frac{172}{3}\)
x + 99 = 172/3 => x =-125/3
y - 98 = - 344/3 => y = - 50 /3
z+ 97 = 172 => z = 75
28 tháng 9 2017
a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)
Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
Vậy x = -5
b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)
\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)
\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
\(\Rightarrow x+102=0\Rightarrow x=-102\)
Vậy x = -102
c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3
= x - x + 2 - 3
= -1
mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0
d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)
\(\Rightarrow x\ge\frac{-7}{3}\)
Vậy \(x\ge\frac{-7}{3}\)
24 tháng 6 2018
\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
Nên \(x+100=0\)
\(\Rightarrow\)\(x=-100\)
Vậy \(x=-100\)
Chúc bạn học tốt ~
24 tháng 6 2018
\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow\)\(x+1=2009\)
\(\Leftrightarrow\)\(x=2009-1\)
\(\Leftrightarrow\)\(x=2008\)
Vậy \(x=2008\)
Chúc bạn học tốt ~
NV
20 tháng 7 2016
a/ \(\frac{5x-4}{3-2x}=\frac{7+4x}{x+2}\) (ĐK: \(x\ne\frac{3}{2};x\ne-2\))
\(\Rightarrow\left(x+2\right)\left(5x-4\right)=\left(7+4x\right)\left(3-2x\right)\)
\(\Rightarrow5x^2-4x+10x-8=21-14x+12x-8x^2\)
\(\Rightarrow13x^2+8x-29=0\)
\(\Rightarrow13\left(x^2+\frac{8}{13}x-\frac{29}{13}\right)=0\)
\(\Rightarrow13\left[x^2+2.\frac{4}{13}.x+\left(\frac{4}{13}\right)^2-\left(\frac{4}{13}\right)^2-\frac{29}{13}\right]=0\)
\(\Rightarrow13\left[\left(x+\frac{4}{13}\right)^2-\frac{393}{169}\right]=0\)
\(\Rightarrow13\left(x+\frac{4}{13}\right)^2-\frac{393}{13}=0\)
\(\Rightarrow\left(x+\frac{4}{13}\right)^2=\frac{393}{169}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{13}=\sqrt{\frac{393}{169}}=\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4+\sqrt{393}}{13}\\x+\frac{4}{3}=-\sqrt{\frac{393}{169}}=-\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4-\sqrt{393}}{13}\end{cases}}\)
Vậy biểu thức có 2 nghiệm \(x=\left\{\frac{-4+\sqrt{393}}{13};\frac{-4-\sqrt{393}}{13}\right\}\)
b/ \(\frac{x-1}{99}+\frac{x-2}{98}-\frac{x-3}{97}-\frac{x-4}{96}=0\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1-\left(\frac{x-3}{97}-1\right)-\left(\frac{x-4}{96}-1\right)=0\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
=> x - 100 = 0 => x = 100
Vậy x = 100
\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=0\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\text{ (do }\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\ne0\text{)}\)
\(\Leftrightarrow x=-100\)
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