Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
do vế trái luôn luôn lớn hơn hoặc =0
=> vế phải cx luôn luôn lớn hơn hoặc =0
=> bỏ giá trị tuyệt đối =100x
có 99x + ........... = 100x
trừ là ra nha bn
ta có:
|x+1/1.2|+|x+1/2.3|+...+|x+1/99.100|=100x
=>|x+1/1.2+x+1/2.3+...+x+1/99.100|=100x
<=>|(x+x+x+...+x)+1/1.2+1/2.3+....1/99.100|=100x
<=>|x.99+1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100|=100x
<=>|x.99+1-1/100|=100x
<=>|99x+99/100|=100x
Có 2 trường hợp
TH1
99x+99/100=100x
=>100x-99x=99/100
<=>x=99/100
=>x=99/100
TH2:
99x+99/100=-100x
-100x-99x=99/100
<=>-199x=99/100
<=>x=99/-19900( loại vì |99x+99/100| là số dương nên 100x là số dương mà x là sô âm nên 100x là số âm)
\(A=\frac{1}{1.2}-x+\frac{1}{2.3}-x+...+\frac{1}{100.101}-x+100x\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}-100x+100x\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=2010\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)\(\Rightarrow\dfrac{-1}{x+3}=\dfrac{1}{2010}\)
\(\Rightarrow x+3=-2010\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=\dfrac{1}{2010}\)
=> \(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)
=> \(-\dfrac{1}{x+3}=\dfrac{1}{2010}\)
=> \(-2010=x+3\)
=> \(x=-2013\)
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)