Cho f(x)=ax^2+bx+c. Biet f(0), f(1), f(2) la so nguyen. Chung minh f(x) luon nhan gia tri nguyen voi moi x nguyen
cho f(x) =ax*2+bx+c biet f(1) .f(2) .f(0) nguyen .chung minh da thuc f(x) nguyen voi moi x
Cho y=f(x)=ax^3+bx^2+cx+d. Biet hamso nhan gia tri nguyen voi moi x nguyen. CM 6a, 2b, a+b+c la cac so nguyen.
CM dieu nguoc lai.
Cho f(x)=ax3+bx2+cx+d. CMR neu 6a, 2b, a+b+c va d la cac so nguyen to thi f(x) co gia tri nguyen voi moi so nguyen x
cho da thuc f(x)=ax^2+bx+c voi a,b,c la cac so thuc . Biet rang f(0), f(1), f(2) co gia tri nguyen . cmr : 2a, 2b cung co gt nguyen
chof(x)=ax^2+bx+cvoi a b c là các số hữu tỉ thỏa mãn 13a+b+2c=0 cmr f(-2)xf(3),nho hon bang 0
Toan lop 7 ma sao kho the?!!!!! Minh bo tay!
cho da thuc f( x) = x4+ 2x3 -x - 2
a, phan tich f(x) thanh nhan tu
b, chung minh f(x) chia het cho 6 voi moi x la so nguyen
a)\(f\left(x\right)=x^4+2x^3-x-2\)
\(=x^4+2x^3+x^2-x^2-x-2\)
\(=\left(x^2+x\right)^2-\left(x^2+x\right)-2\)
Đặt \(x^2+x=t\) ta có:
\(=t^2-t-2\)\(=\left(t-2\right)\left(t+1\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)\)
Cho da thuc f(x)=ax2+bx+c ; a,b,c la cac so nguyen Chung minh rang khong xay ra dong thoi f(2016)=2017; f(2018)=2018
cho da thuc q(x)=ax^2 +bx +c .biet Q(1),Q(-1),Q(0) la so nguyen.cmr voi moi x nguyen thi Q(x) nguyen
Cho da thuc f(x)= ax^2+bx+c bang 0 voi moi gia tri cua x chung minh a=b=c=0
cho ham so f(x) xac dinh voi moi x
Biet vs moi gia tri tuong ung cua x, ta luon co :f(x)+2f(1/x)=x2.Tinh f(4)
\(f\left(4\right)+2f\left(\frac{1}{4}\right)=4^2=16\)(1)
\(f\left(\frac{1}{4}\right)+2f\left(\frac{1}{\frac{1}{4}}\right)=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow f\left(\frac{1}{4}\right)+2f\left(4\right)=\frac{1}{16}\Rightarrow2f\left(\frac{1}{4}\right)+4f\left(4\right)=\frac{1}{8}\)(2)
Từ (1) và (2), ta được:
\(2f\left(\frac{1}{4}\right)+4f\left(4\right)-f\left(4\right)-2f\left(\frac{1}{4}\right)=\frac{1}{8}-16\)
\(\Rightarrow3f\left(4\right)=\frac{-127}{8}\Rightarrow f\left(4\right)=\frac{-127}{24}\)