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 \(f\left(x\right)=10x\)

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

Xét hàm \(g\left(x\right)=f\left(x\right)-10x\)

\(\Rightarrow g\left(1\right)=f\left(1\right)-10.1=10-10=0\)

Tương tự \(g\left(2\right)=0\) ; \(g\left(3\right)=0\)

\(\Rightarrow g\left(x\right)\) luôn có 3 nghiệm \(x=\left\{1;2;3\right\}\)

\(\Rightarrow g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)\) với a là số thực bất kì

\(\Rightarrow f\left(x\right)=g\left(x\right)+10x=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)+10x\)

\(\Rightarrow f\left(12\right)=990\left(12-a\right)+120=12000-990a\)

\(f\left(-8\right)=-990\left(-8-a\right)-80=7840+990a\)

\(\Rightarrow\frac{f\left(12\right)+f\left(-8\right)}{10}+15=\frac{12000-990a+7840+990a}{10}+15=1999\)

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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Xét đa thức Q(x) = P(x) - 10x ,ta có:

Q(1) = P(1) - 10 = 10 - 10 = 0

Q(2) = P(2) - 20 = 20 - 20 = 0

Q(3) = P(3) - 30 = 30 - 30 = 0

=> x = 1 ; x = 2 ; x = 3 là 3 nghiệm của đa thức Q(x), do đó \(Q\left(x\right)⋮\left(x-1\right)\left(x-2\right)\left(x-3\right)\).

=> Q(x) có dạng : 

Q(x) = (x - 1)(x - 2)(x - 3)(x - a)                       \(\left(a\inℚ\right)\)

Khi đó: P(x) = (x - 1)(x - 2)(x - 3)(x - a) + 10x

Ta có: P(12) = 11.10.9.(12 - a) + 120

           P(-8) = -9.(-10).(-11)(-8 - a) - 80

=> P(12) + P(-8) = 11.1019.(12 - a + 8 + a) + 40 

                           = 11.10.9.20 + 40  = 19840

Vậy P(12) + P(-8) = 19840

cái này có trong nâng cao chuyên đề thì phải, nâng cao chuyên đề 8 ấy, e mở ra tham khảo nhá, t nhác vt 

hình như bài 98 thì phải phần đa thức ý

Nâng cao chuyên đề toán 8 đại nhé 

Đặt \(g(x)=10x\).

Ta có \(g\left(1\right)=10=f\left(1\right);g\left(2\right)=20=f\left(2\right);g\left(3\right)=30=f\left(3\right)\).

Từ đó \(\left\{{}\begin{matrix}f\left(1\right)-g\left(1\right)=0\\f\left(2\right)-g\left(2\right)=0\\f\left(3\right)-g\left(3\right)=0\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=Q\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\).

\(\Rightarrow f\left(x\right)=10x+Q\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow f\left(8\right)+f\left(-4\right)=80+Q\left(x\right).7.6.5+\left(-40\right)+Q\left(x\right).\left(-5\right).\left(-6\right).\left(-7\right)=80-50=40\).