Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(G\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-k\right)\)
\(\Rightarrow\hept{\begin{cases}G\left(12\right)=\left(12-1\right)\left(12-2\right)\left(12-3\right)\left(12-k\right)=11880-990k\\G\left(-8\right)=\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)\left(-8-k\right)=7920+990k\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}f\left(12\right)=G\left(12\right)+12\times10=12000-990k\\f\left(-8\right)=G\left(-8\right)+10\times\left(-8\right)=7840+990k\end{cases}}\)
\(\Rightarrow f\left(12\right)+f\left(-8\right)=12000-990k+7840+990k=19840\)
\(\Rightarrow P=\frac{19840}{10}+25=2009\)
Câu này bn lập hpt tìm a,b,c rồi thay 100 và -96 vô tính.
Mk chỉ gợi ý thôi bn tự làm nhé! ^^
\(2,\\ PT\Leftrightarrow6x^2+9y^2-\left(x^2+y^2\right)=20412\\ \text{Mà }20412⋮3;6x^2+9y^2⋮3\\ \Leftrightarrow x^2+y^2⋮3\Leftrightarrow x^2⋮3;y^2⋮3\Leftrightarrow x⋮3;y⋮3\)
Đặt \(\left\{{}\begin{matrix}x=3a\\y=3b\end{matrix}\right.\left(a,b\in Z\right)\Leftrightarrow5\left(3a\right)^2+8\left(3b\right)^2=20412\)
\(\Leftrightarrow9\left(5a^2+8b^2\right)=20412\\ \Leftrightarrow5a^2+8b^2=2268\)
Mà \(2268⋮3\Leftrightarrow5a^2+8b^2⋮3\Leftrightarrow a^2⋮3;b^2⋮3\Leftrightarrow a⋮3;b⋮3\)
Đặt \(\left\{{}\begin{matrix}a=3c\\b=3d\end{matrix}\right.\left(c,d\in Z\right)\Leftrightarrow9\left(5c^2+8d^2\right)=2268\Leftrightarrow5c^2+8d^2=252\)
Mà \(252⋮3\Leftrightarrow5c^2+8d^2⋮3\Leftrightarrow c^2⋮3;d^2⋮3\Leftrightarrow c⋮3;d⋮3\)
Đặt \(\left\{{}\begin{matrix}c=3k\\d=3q\end{matrix}\right.\left(k,q\in Z\right)\Leftrightarrow9\left(5k^2+8q^2\right)=252\Leftrightarrow5k^2+8q^2=28\)
\(\Leftrightarrow5k^2=28-8q^2\ge0\Leftrightarrow q^2\le\dfrac{28}{8}=3,5\\ \text{Mà }q\in Z\\ \Leftrightarrow-3\le q^2\le3\Leftrightarrow-1\le q\le1\)
\(\forall q=0\Leftrightarrow k^2=\dfrac{28}{5}\left(ktm\right)\\ \forall q=\pm1\Leftrightarrow k=\pm2\\ \Leftrightarrow\left(c;d\right)=\left(6;3\right);\left(-6;-3\right);\left(-6;3\right);\left(6;-3\right)\\ \Leftrightarrow\left(a;b\right)=\left(18;9\right)\left(-18;-9\right);\left(-18;9\right);\left(18;-9\right)\\ \Leftrightarrow\left(x;y\right)=\left(54;27\right);\left(-54;-27\right);\left(54;-27\right);\left(-54;27\right)\)
Ta có:
\(P\left(1\right)=7=7.1^2\); \(P\left(2\right)=28=7.2^2\); \(P\left(3\right)=63=7.3^2\)
\(\Rightarrow\)Đặt \(g\left(x\right)=7x^2\).
Đặt \(Q\left(x\right)=P\left(x\right)-g\left(x\right)\).
Ta có:
\(Q\left(1\right)=Q\left(2\right)=Q\left(3\right)=0\)
\(\Rightarrow x=1;x=2;x=3\)là các nghiệm của đa thức Q(x)
\(\Rightarrow Q\left(x\right)⋮\left(x-1\right);\left(x-2\right);\left(x-3\right)\)
Do Q(x) là đa thức bậc 4 có hệ số cao nhất bằng 1 nên
\(Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right).\)
\(\Rightarrow P\left(x\right)=Q\left(x\right)+g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)+7x^2\)
Ta có:
\(P\left(100\right)=\left(100-1\right)\left(100-2\right)\left(100-3\right)\left(100-m\right)+7.100^2\)
\(=99.98.97\left(100-m\right)+7.100^2==97.98.99.100-97.98.99m+7.100^2\)
\(P\left(-96\right)=\left(-96-1\right)\left(-96-2\right)\left(-96-3\right)\left(-96-m\right)+7.\left(-96\right)^2\)
\(=\left(-97\right).\left(-98\right).\left(-99\right).\left(-96-m\right)+7.96^2\)
\(=\left(-96\right).\left(-97\right).\left(-98\right).\left(-99\right)-\left(-97\right).\left(-98\right).\left(-99\right).m+7.96^2\)
\(=96.97.98.99+97.98.99m+7.96^2\)
\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)
\(=\frac{97.98.99.100-97.98.99m+7.100^2+96.97.98.99+97.98.99m+7.96^2}{8}\)
\(=\frac{97.98.99\left(100+96\right)+7.\left(100^2+96^2\right)}{8}=112244867\)
Ta có:
\(P\left(1\right)=7=7.1^2\); \(P\left(2\right)=28=7.2^2\); \(P\left(3\right)=63=7.3^2\)
\(\Rightarrow\)Đặt \(g\left(x\right)=7x^2\).
Đặt \(Q\left(x\right)=P\left(x\right)-g\left(x\right)\).
Ta có:
\(Q\left(1\right)=Q\left(2\right)=Q\left(3\right)=0\)
\(\Rightarrow x=1;x=2;x=3\)là các nghiệm của đa thức Q(x)
\(\Rightarrow Q\left(x\right)⋮\left(x-1\right);\left(x-2\right);\left(x-3\right)\)
Do Q(x) là đa thức bậc 4 có hệ số cao nhất bằng 1 nên
\(Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right).\)
\(\Rightarrow P\left(x\right)=Q\left(x\right)+g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)+7x^2\)
Ta có:
\(P\left(100\right)=\left(100-1\right)\left(100-2\right)\left(100-3\right)\left(100-m\right)+7.100^2\)
\(=99.98.97\left(100-m\right)+7.100^2==97.98.99.100-97.98.99m+7.100^2\)
\(P\left(-96\right)=\left(-96-1\right)\left(-96-2\right)\left(-96-3\right)\left(-96-m\right)+7.\left(-96\right)^2\)
\(=\left(-97\right).\left(-98\right).\left(-99\right).\left(-96-m\right)+7.96^2\)
\(=\left(-96\right).\left(-97\right).\left(-98\right).\left(-99\right)-\left(-97\right).\left(-98\right).\left(-99\right).m+7.96^2\)
\(=96.97.98.99+97.98.99m+7.96^2\)
\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)
\(=\frac{97.98.99.100-97.98.99m+7.100^2+96.97.98.99+97.98.99m+7.96^2}{8}\)
\(=\frac{97.98.99\left(100+96\right)+7.\left(100^2+96^2\right)}{8}=112244867\)
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 \(\frac{P\left(12\right)+P\left(-8\right)}{10}+25=\frac{990n+11880-990n+7920}{10}+25=\frac{19800}{10}+25=2005\)