Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
\(A=\left(\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}+\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}\right):\dfrac{x-1}{x^3}\)
\(=\dfrac{x^2+3}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)^2}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+\right)\left(x+3\right)}+...+\frac{1}{\left(x+2015\right)\left(x+2016\right)}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+2015}-\frac{1}{x+2016}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+2016}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+2016}-\frac{1}{x+2016}=0\)
\(\frac{1}{x}-\frac{2x}{x+2016}=0\)
\(\frac{x+2016}{x\left(x+2016\right)}-\frac{2x}{x\left(x+2016\right)}=0\)
\(\frac{x+2016-2x}{x\left(x+2016\right)}=0\Leftrightarrow2016-x=0\Leftrightarrow x=2016\)
