Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left|-7\right|+8=7+8=15\)
b) \(\left|-7,5\right|+\left|-2,5\right|=7,5+2,5=10\)
c) \(-\left|3,5\right|+\left|5,5\right|-\left|6\right|=-3,5+5,5-6=-4\)
d) \(\left|11,4-3,4\right|+\left|12,4-15,5\right|=\left|8\right|+\left|-3,1\right|=8+3,1=11,1\)
a)|-7|+8=7+8=15
b)|-7,5|+|-2,5|=7,5+2,5=10
c)-|3,5|+|5,5|-|-6|=-3,5+5,5-6=-4
d)|11,4-3,4|+|12,4-15,5|=8+3,1=11,1
a) \(\left|x+4\right|-2=0\)
\(\Leftrightarrow\left|x+4\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=2\\x+4=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=-6\end{cases}}\)
b) Sai đề.
c) \(\left|x-1\right|=2x\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x\\x-1=-2x\end{cases}\left(x\ge0\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\) \(\Leftrightarrow x=\frac{1}{3}\) thỏa mãn
d) \(3x-\left|x+15\right|=\frac{5}{4}\)
\(\Leftrightarrow\left|x+15\right|=3x-\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+15=3x-\frac{5}{4}\\x+15=\frac{5}{4}-3x\end{cases}\left(x\ge\frac{5}{12}\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{65}{8}\\x=-\frac{55}{16}\end{cases}}\) \(\Leftrightarrow x=\frac{65}{8}\) thỏa mãn
a) \(\left|x+4\right|-2=0\)
\(\left|x+4\right|=2\)
\(\Rightarrow x+4=\pm2\)
\(\cdot x+4=2\) \(\cdot x+4=-2\)
\(x=-2\) \(x=-6\)
b) đề sai
c)\(\left|x-1\right|=2x\)
\(\Rightarrow x-1=\pm2x\)
\(\cdot x-1=2x\) \(\cdot x-1=-2x\)
\(x-2x=1\) \(x+2x=1\)
\(\Rightarrow x=-1\) \(\Rightarrow x=\frac{1}{3}\)
d) \(3x-\left|x+15\right|=\frac{5}{4}\)
\(\left|x+15\right|=3x-\frac{5}{4}\)
\(\Rightarrow x+15=\pm\left(3x-\frac{5}{4}\right)\)
\(\cdot x+15=3x-\frac{5}{4}\) \(\cdot x+15=-3x+\frac{5}{4}\)
\(x-3x=-\frac{5}{4}-15\) \(x+3x=\frac{5}{4}-15\)
\(-2x=-\frac{65}{4}\) \(4x=\frac{55}{4}\)
\(x=\frac{65}{8}\) \(x=\frac{55}{8}\)
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
không đâu dễ mà
\(D=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2020.2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
b) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{21.23}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{21.23}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{21}-\frac{1}{23}\right)=\frac{1}{2}\left(1-\frac{1}{23}\right)=\frac{1}{2}.\frac{22}{23}=\frac{11}{23}\)
c) \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}=\frac{1}{99}-\left(\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)=\frac{1}{99}-\left(-\frac{1}{99}+1\right)=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
d) bạn xem lại đề
a)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=\frac{1}{1}-\frac{1}{2021}\)
\(=\frac{2020}{2021}\)
b)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{21\cdot23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\frac{22}{23}\)
\(=\frac{11}{23}\)
c)
\(=\frac{1}{99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=\frac{-97}{99}\)
d)
đề sai hay sao á mong bạn xem ljai ạ