\(x^6-x^4+2x^3+2x^2=y^2\)
\(y^2+y=x^4+x^3+x^2+x=0\left(1\right)\)
\(\Leftrightarrow y\left(y+1\right)=x\left(x^3+x^2+x+1\right)=0\)
Ta có 4 PT
\(x1=0;y1=0\)
\(x2=0;y2=-1\)
\(x3=-1;y3=0\)
\(x4=-1;y4=-1\)
tim x;y biet
x-y=xy-1
cho x thuoc tap so nguyen biet
x^3 -x^2 -2x^2 +2x chia het cho 6
cho x,y thuoc R khac 0 thoa man 2x^2 + y^2/4 +1/x^2 = 4. tim gtnn gtln cua A= 2008+xy
cho x+y=5 tim max A=x^4+y^4-4(x^3+y^3)-20(x^2+y^2)-2x^2y^2+xy
tim x biet x-y=xy-1
cm x^3 -x^2 -2x^2 +2x chia het cho 6
1)Tìm x,y thuoc Z thoa man dong thoi
x^3+y^3=1 x^7+y^7=x^4+y^4
2)Cho A=y^5 - 5y^3 +4y y thuoc Z
CM nếu y ko chia hết 3 thì A chia hết 360
3)Tìm P(x) bậc 4 thỏa mãn
P(-1)=0 , P(x)-P(x-1)=x*(x+1)*(2x+1) voi x thuoc R
\(\frac{6}{x-3}+\frac{2x^2}{^{x^2}-1}+\frac{6-2x}{x^{3^{ }}-3x^2-x+3}\)
a / Rut gon A
b/ Tim x thuoc Z de A thuoc Z
c/ Tinh gia tri cua A khi x = căn 2
cho P=(2x-1)(x+2)/(2x+1)
Tim x thuoc Z để P thuoc Z
chung to rang bieu thuc khong thuoc vao bien
a 2 (2x+x mũ 2 ) + X mũ 2 (x+2) +(x mũ 3+4X+3)
b z (y-x) + y(z -x) + x (y+z) -2yz+ 10
c 2y (y mũ 2+y +1) -2y mũ 2 (y+1)-2(y+1)
d x (3x +12) - (7x-20) +x mũ 2 (2x-3)-3(2x mũ 2+5)
e3(2x-1)-5(x+3)+(3x-4) -19x
1)Tìm x,y thuoc Z thoa man dong thoi
x^3+y^3=1 x^7+y^7=x^4+y^4
2)Cho A=y^5 - 5y^3 +4y y thuoc Z
CM nếu y ko chia hết 3 thì A chia hết 360
3)Tìm P(x) bậc 4 thỏa mãn
P(-1)=0 , P(x)-P(x-1)=x*(x+1)*(2x+1) voi x thuoc R
