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`1/2-(1/3+3/4)<=x<=1/24-(1/8-1/3)`
`<=>6/12-4/12-9/12<=x<=1/24-3/24+8/24`
`<=>-7/12<=x<=1/4`
`<=>-14/24<=x<=3/12`
`=>-14<=x<=3`
`=>x\in{-14;-13;-12;...;3}` do `x\inZZ`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 1/2 - ( 1/3 + 3/4) <= x <= 1/24 - ( 1/8 - 1/3 )
=> 6/12 - ( 4/12 + 9/12 ) <= x <= 1/24 - ( 3/24 - 8/24 )
=> 6/12 - 13/12 <= x <= 1/24 + 5/24
=> -7/12 <= x <= 3/12
=> -7 <= 12x <= 3
=> x ko tồn tại
1/3– 3/5 + 5/7 –7/9 + 9/11 – 11/13 + 13/15 + 11/13 – 9/11 + 7/9 –5/7 + 3/5 –1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}-\left\{\frac{1}{3}+\frac{3}{4}\right\}\)
\(=\frac{1}{2}-\frac{13}{12}\)
\(=\frac{1}{2}-1+\frac{1}{12}\)
\(=\frac{7}{12}-1\)
\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\)
\(=\frac{1}{2}-\frac{13}{12}\)
\(=\frac{-7}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)=\left(7-x\right)\left(7+x\right)\)
\(\Leftrightarrow3x-1=7-x\)
\(\Leftrightarrow4x=8\Leftrightarrow x=2\)
Vậy x = 2
2, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
Vậy x = -2 hoặc x = 0
3, \(\left(1-2x\right)^2-\left(x+3\right)^2+3\left(x+1\right)\left(1-x\right)=8\)
\(\Leftrightarrow\left(1-2x-x-3\right)\left(1-2x+x+3\right)+3\left(x-x^2+1-x\right)=8\)
\(\Leftrightarrow\left(-2-3x\right)\left(4-x\right)-3x^2+3=8\)
\(\Leftrightarrow-8+2x-12x+3x^2-3x^2=5\)
\(\Leftrightarrow-10x=13\)
\(\Leftrightarrow x=-1,3\)
Vậy x = -1,3
4, \(\left(x-3\right)^2-\left(x+3\right)^2=24\)
\(\Leftrightarrow\left(x-3-x-3\right)\left(x-3+x+3\right)=24\)
\(\Leftrightarrow-6.2x=24\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(K=\left(1-\dfrac{3}{2\cdot4}\right)\left(1-\dfrac{3}{3\cdot5}\right)\cdot...\cdot\left(1-\dfrac{3}{19\cdot21}\right)\)
\(=\dfrac{3^2-1-3}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2-1-3}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{20^2-4}{\left(20-1\right)\left(20+1\right)}\)
\(=\dfrac{\left(3-2\right)\left(3+2\right)}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{\left(4-2\right)\left(4+2\right)}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{18\cdot22}{\left(20-1\right)\left(20+1\right)}\)
\(=\dfrac{1\cdot5}{2\cdot4}\cdot\dfrac{2\cdot6}{3\cdot5}\cdot...\cdot\dfrac{18\cdot22}{19\cdot21}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot21\cdot22}{2\cdot3\cdot4\cdot5\cdot...\cdot19\cdot20\cdot21}=1\cdot22=22\)
![](https://rs.olm.vn/images/avt/0.png?1311)