Lời giải:

Ta có thể viết dạng của $f(x)$ như sau:

\(f(x)=(x-1)(x-2)(x-3)(x-t)+g(x)\)

Trong đó, \(t\) là một số bất kỳ nào đó và \(g(x)\) là đa thức có bậc nhỏ hơn hoặc bằng $3$

Giả sử \(g(x)=mx^3+nx^2+px\)

\(\left\{\begin{matrix} f(1)=g(1)=m+n+p=10\\ f(2)=g(2)=8m+4n+2p=20\\ f(3)=g(3)=27m+9n+3p=30\end{matrix}\right.\)

Giải hệ trên thu được \(m=0,n=0,p=10\)

Như vậy \(f(x)=(x-1)(x-2)(x-3)(x-t)+10x\)

Do đó \(\left\{\begin{matrix} f(12)=990(12-t)+120=12000-990t\\ f(-8)=-990(-8-t)-80=7840+990t\end{matrix}\right.\)

\(\Rightarrow \frac{f(12)+f(-8)}{10}+26=\frac{12000+7840}{10}+26=2010\) (đpcm)

Đặt \(f\left(x\right)=10x\)

Khi đó ta có \(f\left(1\right)=10=P\left(1\right)\), \(f\left(2\right)=20=P\left(2\right)\), \(f\left(3\right)=30=P\left(3\right)\)

Do đó \(P\left(x\right)-f\left(x\right)=g\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow P\left(x\right)=10+g\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

Vì \(P\left(x\right)\)là đa thức bậc 4 mà \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\)là đa thức bậc 3 nên \(g\left(x\right)\)là đa thức bậc 1 hay \(g\left(x\right)=x+n\)

Vậy \(P\left(x\right)=\left(x+n\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)+10\)

\(\Rightarrow P\left(12\right)=\left(12+n\right)\left(12-1\right)\left(12-2\right)\left(12-3\right)=\left(n+12\right).11.10.9=990\left(n+12\right)\)

\(=990n+11880\)

Và \(P\left(-8\right)=\left(-8+n\right)\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)=\left(n-8\right)\left(-9\right)\left(-10\right)\left(-11\right)\)\(=-990\left(n-8\right)=-990n+7920\)

Vậy \(\frac{P\left(12\right)+P\left(-8\right)}{10}+25=\frac{990n+11880-990n+7920}{10}+25=\frac{19800}{10}+25=2005\)

Đặt Q(x)=P(x)-10x. Khi đó Q(1)=Q(2)=Q(3)=0

Vì vậy Q(x) chia hết cho (x-1)(x-2)(x-3). Q(x) là đa thức bậc 4 (do P(x) là đa thức bậc 4) nên Q(x)=(x-1)(x-2)(x-3)(x-r) và 

P(x)=(x-1)(x-2)(x-3)(x-r)+10x

P(12)=1200-990r

P(-8)=7840+990r

Vậy \(\frac{P\left(12\right)+P\left(-8\right)}{10}=1984\)

Ta có:      \(P\left(1\right)=1+a+b+c+d=10\)
                \(P\left(2\right)=16+8a+4b+2c+d=20\)
                \(P\left(3\right)=81+27a+9b+3c+d=30\)
    và        \(P\left(12\right)=20736+1728a+144b+12c+d\)
                 \(P\left(-8\right)=4096-512a+64b-8c+d\)
suy ra   \(P\left(12\right)+P\left(-8\right)=24832+1216a+208b+4c+2d\)

Ta lại có:               \(100.P\left(1\right)-198.P\left(2\right)+100.P\left(3\right)\)     \(=100\left(1+a+b+c+d\right)-198\left(16+8a+4b+2c+d\right)+100\left(81+27a+9b+3c+d\right)\)
\(=100+100a+100b+100c+100d-3168-1584a-792b-396c-198d+8100+2700a+900b+300c+100d\)
\(=5032+1216a+208b+4c+2d\)

Mặt khác:                      \(100.P\left(1\right)-198.P\left(2\right)+100.P\left(3\right)\)
    \(=100\times10-198\times20+100\times30=40\)

Do đó:          \(5032+1216a+208b+4c+2d=40\)
       \(\Rightarrow\)\(1216a+208b+4c+2d=40-5032=-4992\)

Thế  \(1216a+208b+4c+2d=-4992\)  vào \(P\left(12\right)+P\left(-8\right)=24832+1216a+208b+4c+2d\)
ta được:    \(P\left(12\right)+P\left(-8\right)=24832-4992=19840\)

Vậy  \(\frac{P\left(12\right)+P\left(-8\right)}{10}=\frac{19840}{10}=1984\)

trong cac dong sau dong nao cho 3 so tu nhien lien tiep tang dan

Đặt \(f\left(x\right)=P\left(x\right)+3x\)

\(f\left(x\right)=P\left(x\right)+3x\\ \Leftrightarrow\left\{{}\begin{matrix}f\left(-2\right)=0\\f\left(-4\right)=0\\f\left(-6\right)=0\end{matrix}\right.\Leftrightarrow f\left(x\right)=\left(x-m\right)\left(x+2\right)\left(x+4\right)\left(x+6\right)\\ \Leftrightarrow P\left(x\right)=\left(x-m\right)\left(x+2\right)\left(x+4\right)\left(x+6\right)+3x\\ \Leftrightarrow\left\{{}\begin{matrix}P\left(-2\right)=0\\P\left(0\right)=-m\cdot2\cdot4\cdot6+0=-48m\\P\left(-8\right)=\left(-8-m\right)\left(-6\right)\left(-4\right)\left(-2\right)-24=48m+360\end{matrix}\right.\)

Do đó \(A=\dfrac{-48m+48m+360+0}{2020}=\dfrac{360}{2020}=\dfrac{18}{101}\)

Ta có: 

\(P\left(1\right)=a+b+c+d+1\)

\(P\left(2\right)=8a+4b+2c+d+16\)

\(P\left(3\right)=27a+9b+3c+d+81\)

\(\Rightarrow100P\left(1\right)-198P\left(2\right)+100P\left(3\right)\)

\(=100\left(a+b+c+d+1\right)-198\left(8a+4b+2c+d+16\right)+100\left(27a+9b+3c+d+81\right)\)

\(=1216a+208b+4c+2d+5032=100.10-198.20+100.30=40\)

Ta lại có: 

\(f\left(12\right)+f\left(-8\right)=12^4+12^3a+12^2b+12c+d+8^4-8^3a+8^2b-8c+d\)

\(=\left(1216a+208b+4c+2d+5032\right)+19800\)

\(=40+19800=19840\)

\(\Rightarrow P=\frac{19840}{10}+25=2009\)

Đặt \(G\left(x\right)=f\left(x\right)-10x\)\(\Leftrightarrow\hept{f\left(x\right)=G\left(x\right)+10x}\)và \(G\left(x\right)\)có bậc 4 có hệ số cao nhất là 1

Từ đề bài ta có: \(\hept{\begin{cases}G\left(1\right)=f\left(1\right)-10=0\\G\left(2\right)=f\left(2\right)-20=0\\G\left(3\right)=f\left(3\right)-30=0\end{cases}}\)\(\Rightarrow x=1;2;3\)là 3 nghiệm của\(G\left(x\right)\)

\(\Rightarrow G\left(x\right)\)có dạng \(G\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-k\right)\)

\(\Rightarrow\hept{\begin{cases}G\left(12\right)=\left(12-1\right)\left(12-2\right)\left(12-3\right)\left(12-k\right)=11880-990k\\G\left(-8\right)=\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)\left(-8-k\right)=7920+990k\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}f\left(12\right)=G\left(12\right)+12\times10=12000-990k\\f\left(-8\right)=G\left(-8\right)+10\times\left(-8\right)=7840+990k\end{cases}}\)

\(\Rightarrow f\left(12\right)+f\left(-8\right)=12000-990k+7840+990k=19840\)

\(\Rightarrow P=\frac{19840}{10}+25=2009\)

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