mn giúp mik vs ạ<3
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a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\)
= \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)
= \(\dfrac{2}{2}+\dfrac{3}{2}-\dfrac{4}{2}\)
= \(\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
10³ + 2¹⁵
= 1000 + 32768
= 33768
Mà 33768 : 33 = 1023 (dư 9)
Em xem lại đề
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Sửa đề P = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)CM P < 5/16
=> 5P = \(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\)
Lấy 5P trừ P theo vế ta có
5P - P = \(\left(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\right)\)
4P = \(1+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{5^2}-\frac{2}{5^2}\right)+...+\left(\frac{11}{5^{10}}-\frac{10}{5^{10}}\right)-\frac{11}{5^{11}}\)
4P = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
Đặt Q = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
=> 5Q \(=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)
Lấy 5Q trừ Q theo vế ta có
5Q - Q = \(\left(5+1+\frac{1}{5}+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)\)
4Q \(=5-\frac{1}{5^{10}}\)
=> Q\(=\frac{5}{4}-\frac{1}{5^{10}.4}\)
Khi đó 4P = \(\frac{5}{4}-\frac{1}{5^{10}.4}-\frac{11}{5^{11}}\)
=> P = \(\frac{5}{16}-\frac{1}{5^{10}.16}-\frac{11}{5^{11}.4}\)
\(=\frac{5}{16}-\frac{1}{5^{10}}\left(\frac{1}{16}-\frac{11}{5.4}\right)< \)\(\frac{5}{16}\)
Bài làm:
Ta có: \(\frac{1}{5^2}+\frac{2}{5^2}+\frac{3}{5^2}+...+\frac{11}{5^2}\)
\(=\frac{1+2+3+...+11}{5^2}=\frac{\left(1+11\right).11:2}{5^2}=\frac{66}{25}>1>\frac{1}{16}\)
\(\Rightarrow P>\frac{1}{16}\)
=> Đề sai
![](https://rs.olm.vn/images/avt/0.png?1311)
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c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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