tra gút gồ đe=))

Đề HSG Nghệ An ak bạn 

P = \(n^4-1=\left(n^2-1\right)\left(n^2+1\right)=\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)

\(=\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)=\left(n-2\right)\left(n+2\right)\left(n-1\right)\left(n+1\right)+5\left(n-1\right)\left(n+1\right)\)

P \(⋮5\Leftrightarrow Q=\left(n-2\right)\left(n-1\right)\left(n+1\right)\left(n+2\right)⋮5\)

mà n không chia hết cho 5 => có dạng n = 5k + 1 ;5k + 2 ; 5k + 3 ;5k + 4 (k \(\in Z\)) 

Khi n = 5k + 1 => n - 1 \(⋮5\Rightarrow Q⋮5\Rightarrow P⋮5\)

tương tự với n = 5k + 2 ; n = 5k + 3 ; n = 5k + 4 thì Q \(⋮5\Rightarrow P⋮5\)

a: a+c=b-8

=>a-b+c=-8

\(G\left(-1\right)=a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c\)

=a-b+c

=-8

b: G(0)=4

=>\(a\cdot0^2+b\cdot0+c=4\)

=>c=4

=>G(x)=\(a\cdot x^2+bx+4\)

G(1)=9

=>\(a\cdot1^2+b\cdot1+4=9\)

=>a+b=5

G(2)=14

=>\(a\cdot2^2+b\cdot2+4=14\)

=>4a+2b=10

=>2a+b=5

mà a+b=5

nên 2a+b-a-b=5-5=0

=>a=0

a+b=5

=>b=5-a=5-0=5

a) Ta có: \(A\left(x\right)=ax^2+bx+c\)

Thay \(A\left(-1\right)\)  ta được:

\(A\left(-1\right)=a\left(-1\right)^2+b\left(-1\right)+c=a+c-b\)

\(=b-8-b=-8\)

b) \(\left\{{}\begin{matrix}A\left(0\right)=4\\A\left(1\right)=9\\A\left(2\right)=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b+c=9\\4a+2b+c=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\4a+2b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\2a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a=0\\b=5\end{matrix}\right.\)

c) 

Ta có: \(\left\{{}\begin{matrix}A\left(2\right)=4a+2b+c\\A\left(-1\right)=a-b+c\end{matrix}\right.\)

\(\Leftrightarrow A\left(2\right)+A\left(-1\right)=5a+b+2c=0\)

\(\Leftrightarrow A\left(2\right)=-A\left(-1\right)\)

\(\Leftrightarrow A\left(2\right)\times A\left(-1\right)=-\left[A\left(2\right)\right]^2\le0\)

\(\frac{a}{-3}=\frac{b}{4};\frac{b}{2}=\frac{c}{3}=>\frac{a}{-3}=\frac{b}{4}=\frac{2}{6}\)

áp dụng tính chất DTSBN ta có

\(\frac{a}{-3}=\frac{b}{4}=\frac{c}{6}=\frac{a+b+c}{-3+4+6}=\frac{14}{7}=2\)

\(+\frac{a}{-3}=>a=-6\)

\(+\frac{b}{4}=2=>b=8\)

\(+\frac{c}{6}=2=>c=12\)

Ta có;\(\frac{a}{-3}=\frac{b}{4};\frac{b}{2}=\frac{c}{3}\Leftrightarrow\frac{b}{4}=\frac{c}{6}\Rightarrow\frac{a}{-3}=\frac{b}{4}=\frac{c}{6}\)

Áp dụng tính chất dãy tỉ số băng nhau:

 \(\frac{a}{-3}=\frac{b}{4}=\frac{c}{6}=\frac{a+b+c}{-3+4+6}=\frac{14}{7}=2\)

Vậy\(\hept{\begin{cases}a=2\cdot\left(-3\right)=-6\\b=2\cdot4=8\\c=2\cdot6=12\end{cases}}\)

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