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\(\Rightarrow\)\(\frac{2}{6}\)+ \(\frac{2}{12}\)+ \(\frac{2}{20}\)+...+\(\frac{2}{x\left(x+1\right)}\)= \(\frac{2011}{2013}\)
\(\Rightarrow\)\(\frac{2}{2.3}\)+ \(\frac{2}{3.4}\)+ \(\frac{2}{4.5}\)+...+ \(\frac{2}{x\left(x+1\right)}\)= \(\frac{2011}{2013}\)
\(\Rightarrow\)\(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+...+ \(\frac{1}{x}\)- \(\frac{1}{x+1}\)= \(\frac{2011}{2013}\): 2
\(\Rightarrow\)\(\frac{1}{2}\)- \(\frac{1}{x+1}\)= \(\frac{2011}{4026}\)
\(\Rightarrow\)\(\frac{1}{x+1}\)= \(\frac{1}{2}\)- \(\frac{2011}{4026}\)= \(\frac{1}{2013}\)
\(\Rightarrow\)\(x+1=2013\)
b)
\(\left(x-3\right)\left(2y+1\right)=7=1.7=\left(-1\right).\left(-7\right)\)
TH1:
\(\hept{\begin{cases}x-3=1\\2y+1=7\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=3\end{cases}}}\)
TH2:
\(\hept{\begin{cases}x-3=7\\2y+1=1\end{cases}\Rightarrow\hept{\begin{cases}x=10\\y=0\end{cases}}}\)
TH3:
\(\hept{\begin{cases}x-3=-1\\2y+1=-7\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-4\end{cases}}}\)
TH4:
\(\hept{\begin{cases}x-3=-7\\2y+1=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-1\end{cases}}}\)
Vậy....
a, 30(x+2) - 6(x-5) - 24x=100
30x + 60 - 6x + 30 - 24x = 100
24x + 90 - 24x = 100
0x = 10 => Không có giá trị x nào thỏa mãn
\(\frac{-2}{3}\) \(-\) \(\frac{1}{3}\) X \(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(-1\) X \(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(\left(2.x-5\right)\) \(=\frac{3}{2}\) \(:-1\)
\(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(2.x\) \(=\frac{3}{2}\) \(+\) \(5\)
\(2.x\) \(=\frac{7}{2}\)
\(x=\) \(\frac{7}{2}\) \(:2\)
\(x=\frac{7}{4}\)
* Mới lớp 5 nên không chắc, sai thongcam *
#Ninh Nguyễn
\(\frac{-2}{3}-\frac{1}{3}\cdot\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-2}{3}-\frac{3}{2}\)
\(2x-5=\frac{-13}{6}:\frac{1}{3}\)
\(2x=\frac{-13}{2}+5\)
\(x=\frac{-3}{2}:2\)
\(x=\frac{-3}{4}\)
\(A=1+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+...+2^{2019}\)
\(2A-A=\left(2+2^2+...+2^{2019}\right)-\left(1+2^2+2^3+...+2^{2018}\right)\)
\(A=2^{2019}-1\)
\(\Rightarrow A+1=2^{2019}-1+1=2^{2019}\)
\(\Rightarrow A+1\)là một lũy thừa
đpcm
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\(x-\left[-x+\left(x-3\right)\right]-\left[\left(x+3\right)-\left(x-2\right)\right]=0\)
Phá ngoặc, thu gọn được \(x=2\)
Vậy số đối của x là \(-x=2\Leftrightarrow x=-2\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot.....\cdot\left(1-\frac{1}{2020}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{2019}{2020}\)
\(=\frac{1\cdot2\cdot3\cdot.....\cdot2019}{2\cdot3\cdot4\cdot....\cdot2020}=\frac{1}{2020}\)
\(2\left|x-1\right|+3\left(x+2\right)=3^2\)
\(2\left|x-1\right|+3x+6=9\)
\(2\left|x-1\right|=9-3x-6\)
\(2\left|x-1\right|=3-3x\)
\(\left|x-1\right|=\frac{3-3x}{2}\)
\(\Rightarrow\orbr{\begin{cases}x-1=\frac{3-3x}{2}\\x-1=-\frac{3-3x}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3-3x}{2}+\frac{2}{2}\\x=\frac{-3+3x}{2}+\frac{2}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3-3x+2}{2}\\x=\frac{-3+3x+2}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5-3x}{2}\\x=\frac{-1+3x}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=5-3x\\2x=-1+3x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x=5\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)