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do vế trái lớn hơn hoặc bằng 0
=> 100.x lớn hơn hoặc bằng 0
=> x lớn hơn hoặc bằng 0
=> vế trái
=\(x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}\)
=>101x+\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)
=>x=\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
bạn tự tính vế phải nha
a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)
\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)
\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)
\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)
=> x = 4
b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)
\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)
\(\Rightarrow100x-99x=1-\frac{1}{100}\)
\(\Rightarrow x=\frac{99}{100}\)
phúc hơi phức tạp
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(x=2009-1\)
\(x=2008\)
Vậy \(x=2008\)
Tự làm bước biến đổi nhé tui lm lẹ luôn =v
\(\frac{1}{1}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1}{x+1}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(=>x=2008\)
Vậy x = 2008
Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x
\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)
\(\Rightarrow100x\ge0\)
\(\Rightarrow x\ge0\)
Từ điều kiện trên ta có :
\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)
\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(50x=1-\frac{1}{100}\)
\(50x=\frac{99}{100}\)
\(x=\frac{99}{5000}\)
Do \(\left|a\right|\ge0\forall a\) nên:
\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)
\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)
Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)
\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)
\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)
các giá trị tuyệt đối trên có tổng lớn hơn hoặc bằng 0(>=0)
=>100x>=0
=>x>=0 =>x+1/(1.2) >0 ;x+1/(2.3)>0;x+1/(3.4);.....;x+1/(99.100)>0
=> ta có thể phá dấu giá trị tuyệt đối
=>100x=x+x+...+x(có 99. x)+(1/(1.2)+1/(2.3)+..+1/(99.100))
=>100x=99x+99/100
=>x=99/100
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)\)
\(\Rightarrow\left(\frac{x+1}{2009}+\frac{2009}{2009}\right)+\left(\frac{x+2}{2008}+\frac{2008}{2008}\right)=\left(\frac{x+3}{2007}+\frac{2007}{2007}\right)+\left(\frac{x+4}{2006}\frac{2006}{2006}\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2007}+\frac{x+2010}{2006}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2007}-\frac{x+2010}{2006}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
=>x+2010=0
=>x=-2010
Vậy x = -2010
\(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2006.2007}=\frac{2006}{2007}\)
\(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+\frac{x}{3}-\frac{x}{4}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x}{2007}-\frac{x}{2007}=\frac{2006}{2007}\)
\(2007x-x=2006\)
\(2006x=2006\)
\(x=1\)
theo mình x =1 không biết đúng không