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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
\(x+2=41\)
\(x=41-2\)
\(x=39\)
Tìm x
a) \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{x\times\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+1\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{40}{41}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{\left(x+2\right)}=\frac{40}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Rightarrow x+2=41\)
\(\Rightarrow x=41-2\)
\(\Rightarrow x=39\)
Vậy x = 39
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x\cdot\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x\cdot\left(x+2\right)}\right)=\frac{20}{41}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
\(\Rightarrow x+2=41\Rightarrow x=39\)
Cần bổ sung điều kiện \(x;y\inℤ\)
a) \(\left(x-7\right)\left(xy+1\right)=9=1\cdot9=3\cdot3=\left(-1\right)\cdot\left(-9\right)=\left(-3\right)\cdot\left(-3\right)\)
Xét bảng :
x-7 | 1 | 9 | -1 | -9 | 3 | -3 |
xy+1 | 9 | 1 | -9 | -1 | 3 | -3 |
x | 8 | 16 | 6 | -2 | 10 | 4 |
y | 1 | 0 | -1,(6) | 1 | 0,2 | -1 |
Vì x,y thuộc Z nên ta có (x;y)={(8;1),(16;0),(-2;1),(4;-1)
b) \(\left(x+5\right)\left(3x-12\right)>0\)
TH1 : \(\hept{\begin{cases}x+5>0\\3x-12>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-5\\x>4\end{cases}\Leftrightarrow x>4}}\)
TH2 : \(\hept{\begin{cases}x+5< 0\\3x-12< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -5\\x< 4\end{cases}\Leftrightarrow x< -5}}\)
Vậy....
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
(7.x + 1) : 2 - 23 = 2
=> (7.x + 1) : 2 = 2 + 23
=> (7.x + 1) : 2 = 25
=> 7.x + 1 = 25 . 2
=> 7.x + 1 = 50
=> 7.x = 50 - 1
=> 7.x = 49
=> x = 49 : 7 = 7
(7.x + 1) : 2 - 23 = 2
=> (7.x + 1) : 2 = 2 + 23
=> 7.x + 1 = 25 . 2
=> 7.x = 50 - 1
=> 7.x = 49
x = 49 : 7 = 7
Vậy.............
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