1/4-(2x+1/2)mũ 2 =0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Số hạng thứ 20 (n=20) là
\(\left(20-1\right).4=76\)
\(A=1-5+9-13+17-21+...+76\)
\(A=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(A=\left(-4\right).38=-152\)
b) Số hạng thứ n là:
\(\left(n-1\right).4\)
\(\)\(A=1-5+9-13+17-21+...+\left(n-1\right).4\)
\(A=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\) ((n-1).2 số -4)
\(A=\left(-4\right).\left(n-1\right).2=-8\left(n-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5}{7}:x+1=\dfrac{-15}{30}\)
\(\dfrac{5}{7}:x=\dfrac{-15}{30}-1\)
\(\dfrac{5}{7}:x=\dfrac{-45}{30}\)
\(x=\dfrac{5}{7}:\dfrac{-45}{30}\)
\(x=\dfrac{-10}{21}\)
\(\dfrac{5}{7}:x+1=-\dfrac{15}{30}\)
\(\Rightarrow\dfrac{5}{7}:x+1=-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{1}{2}-1\)
\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{5}{7}:-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{10}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{9}{3}\)
\(\Rightarrow x-\dfrac{1}{2}=3\)
\(\Rightarrow x=3+\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{7}{2}\)
\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\\ \\ \\ \Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}=\dfrac{9}{3}=3\\ \\ \\ \Rightarrow x=3+\dfrac{1}{2}=3\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{9}{2}-2x\right)\cdot1\dfrac{4}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{11}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\dfrac{9}{2}-2x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}\cdot\dfrac{7}{11}=\dfrac{1}{2}\\ \\ \\ \Rightarrow2x=\dfrac{9}{2}-\dfrac{1}{2}=\dfrac{8}{2}=4\\ \\ \\ \Rightarrow x=4:2=2\)
(\(\dfrac{9}{2}\) - 2\(x\)) \(\times\) 1\(\dfrac{4}{7}\) = \(\dfrac{11}{14}\)
( \(\dfrac{9}{2}\) - 2\(x\)) \(\times\) \(\dfrac{11}{7}\) = \(\dfrac{11}{14}\)
\(\dfrac{9}{2}\) - 2\(x\) = \(\dfrac{11}{14}\) : \(\dfrac{11}{7}\)
\(\dfrac{9}{2}\) - 2\(x\) = \(\dfrac{1}{2}\)
2\(x\) = \(\dfrac{9}{2}\) - \(\dfrac{1}{2}\)
2\(x\) = 4
\(x\) = 4 : 2
\(x\) = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề ta có y = ax, x = bz. Suy ra y = ax = abz.
Vậy y tỉ lệ thuận với z theo hệ số tỉ lệ ab
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\dfrac{17}{14}:\left(-\dfrac{34}{7}\right)+\dfrac{10}{3}\left(\dfrac{1}{5}-\dfrac{3}{4}\right)-\dfrac{6}{5}\left(\dfrac{1}{3}-\dfrac{5}{6}\right)\)
\(=\dfrac{17}{14}.\dfrac{7}{34}+\dfrac{10}{3}.\left(-\dfrac{11}{20}\right)-\dfrac{6}{5}.\left(-\dfrac{9}{18}\right)\)
\(=\dfrac{1}{4}-\dfrac{11}{6}+\dfrac{9}{15}\)
\(=\dfrac{15-110+36}{60}\)
\(=-\dfrac{59}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy chỉ có \(n=3\) thỏa mãn đẳng thức \(\left(n+5\right)^2=64\left(n-2\right)^3\)vì
- \(\left(n+5\right)^2\) là 1 số chính phương
- \(64\) là 1 số chính phương
- \(\left(n-2\right)^3\) không phải số chính phương
- \(\)\(\left(n+5\right)^2< 64\left(n-2\right)^3,\forall n>3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)
\(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
=> \(\left(2x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)
=> \(2x+\dfrac{1}{2}=\pm\dfrac{1}{2}\)
TH1:
\(2x+\dfrac{1}{2}=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-\dfrac{1}{2}=0\)
\(x=0\)
TH2:
\(2x+\dfrac{1}{2}=-\dfrac{1}{2}\)
\(2x=-\dfrac{1}{2}-\dfrac{1}{2}\)
\(2x=-1\)
\(x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{0;\dfrac{-1}{2}\right\}\)