tìm và phân tích tác dụng biện pháp tu từ trong bài người mẹ vườn cau
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a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
Từ "lạc trôi" có nghĩa là gì trong câu:
"Mây bềnh bồng lạc trôi/mượt mà như tuổi ngọc."
Ta có:
`10^9 - 1`
`= (10^3)^3 - 1^3`
`= 1000^3 - 1^3`
`= (1000 - 1)(1000^2 + 1000 . 1 + 1^2)`
`= 999 . (1000^2 + 1000 + 1) \vdots999 (đpcm)`
Vậy: `10^9 - 1 \vdots 999`
Cách 1:
(a\(x^2\) + b\(x\) + c).(\(x+3\))
= a\(x^3\) + 3a\(x^2\) + b\(x^2\) + 3b\(x\) + c\(x\) + 3c
= a\(x^3\) + (3a\(x^2\) + b\(x^2\)) + (3b\(x\) + c\(x\)) + 3c
= a\(x^3\) + \(x^2\).(3a + b) + \(x\).(3b + c) + 3c
a\(x^3\) + (3a + b)\(x^2\) + (3b + c)\(x\) + 3c = \(x^3\) + 2\(x^2\) - 3\(x\)
⇔ \(\left\{{}\begin{matrix}a=1\\3a+b=2\\3b+c=-3\\3c=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}a=1\\3+b=2\\3b+c=-3\\c=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}a=1\\b=2-3\\3b=-3\\c=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}a=1\\b=-1\\b=-1\\c=0\end{matrix}\right.\)
Vậy (a; b; c) = (1; -1; 0)
Cách hai ta có:
\(x^3\) + 2\(x^2\) - 3\(x\) = (\(x^3\) + 3\(x^2\)) - (\(x^2\) + 3\(x\))
\(x^3\) + 2\(x^2\) - 3\(x\) = \(x^2\).(\(x+3\)) - \(x\).(\(x+3\))
\(x^3\) + 2\(x^2\) - 3\(x\) = (\(x+3\)).(\(x^2\) - \(x\))
⇒ (a\(x^2\) + b\(x\) + c).(\(x\) + 3) = (\(x+3\)).(\(x^2\) - \(x\))
⇔ a\(x^2\) + b\(x\) + c = \(x^2\) - \(x\)
⇒ \(\left\{{}\begin{matrix}a=1\\b=-1\\c=0\end{matrix}\right.\)
Vậy (a; b; c) = (1; -1; 0)