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Có:
a1+a2=a3+a4=...=a2015+a1=1
=>a1+a2+a3+a4+...+a2014+a2015=1007+a2015
Mà 1007+a2015=0
=>a2015=-1007.
=>a1=1--1007
a1=1008.
Chúc học tốt^^
Có:
a1+a2=a3+a4=...=a2015+a1=1
=>a1+a2+a3+a4+...+a2014+a2015=1007+a2015
Mà 1007+a2015=0
=>a2015=-1007.
=>a1=1--1007
a1=1008.
Chúc học tốt^^
Ta có:
\(\begin{cases}a_2^2=a_1.a_3\\a_3^2=a_2.a_4\end{cases}\)\(\Rightarrow\begin{cases}\frac{a_2}{a_3}=\frac{a_1}{a_2}\\\frac{a_3}{a_4}=\frac{a_2}{a_3}\end{cases}\)\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(\Rightarrow\frac{a_1^3}{a_2^3}=\frac{a_2^3}{a_3^3}=\frac{a_3^3}{a_4^3}=\frac{a_1}{a_2}.\frac{a_2}{a_3}=\frac{a_3}{a_4}=\frac{a_1}{a_4}\left(1\right)\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a_1^3}{a_2^3}=\frac{a_2^3}{a_3^3}=\frac{a_3^3}{a_4^3}=\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}=\frac{a_1}{a_4}\left(đpcm\right)\)
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(a2 - 1)(a2 - 4)(a2 - 7)(a2 - 10) < 0
=> (a\(^2\)- 1 ) = 0 => a\(^2\)=1 => a = +-1
=> (a\(^2\)- 4 ) = 0 => a\(^2\)= 4 => a = +-2
=> (a\(^2\)- 7 ) = 0 => a\(^2\)= 7 => a = rỗng ( vì a nguyên )
=> (a\(^2\)- 10 ) = 0 => a\(^2\)= 10 => a = rỗng ( vì a nguyên )
Vậy, ..............
Cô hướng dẫn em lập bảng xét dấu:
Từ bảng xét dấu trên ta có :
\(\left(a^2-1\right)\left(a^2-4\right)\left(a^2-7\right)\left(a^2-10\right)< 0\)
\(\Leftrightarrow-\sqrt{10}< a< -\sqrt{7}\) hoặc -2 < a < -1 hoặc 1 < a < 2 hoặc \(\Leftrightarrow\sqrt{7}< a< \sqrt{10}\)
Do a nguyên nên \(\orbr{\begin{cases}a=-3\\a=3\end{cases}}\)