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Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 M=x3+x2y−2x2−xy−y2+3y+x−1M=x3+x2y−2x2−xy−y2+3y+x−1
M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)
M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1
M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1
M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1
M=x2.0+y.0+0+1M=x2.0+y.0+0+1
M=1M=1
N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2
N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)
N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2
N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2
N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2
N=x2.0−xy.0+2.0+2N=x2.0−xy.0+2.0+2
N=2N=2
P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3
P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3
P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3
P=x3.0+x2y.0−x.0+3P=x3.0+x2y.0−x.0+3
P=3
Lời giải:
$x+y-2=0\Rightarrow x+y=2$
a)
$B=x^4+2x^3y-2x^3+x^2y^2-2x^2y-x(x+y)+2x+3$
$=x^3(x+y)+x^3y-2x^3+x^2y^2-2x^2y-2x+2x+3$
$=2x^3+x^3y-2x^3+x^2y^2-2x^2y+3$
$=x^3y+x^2y^2-2x^2y+3$
$=xy(x^2+xy-2x)+3=xy[x(x+y)-2x]+3=xy(2x-2x)+3=3$
b)
$C=x^3+x^2y-2x^2-xy+y^2-3y-x+5$
$=x^2(x+y)-2x^2-xy+y^2-3(y+x)+2x+5$
$=2x^2-2x^2-xy+y^2-6+2x+5$
$=-xy+y^2+2x-1$
$=y(x+y)+2x-1-2xy=2y+2x-1-2x=2(x+y)-1-2x=3-2x$ (không tính cụ thể được giá trị- bạn xem lại đề)
c)
$D=2x^4+3x^2y^2+y^4+y^2$
$=(x^4+2x^2y^2+y^4)+x^4+x^2y^2+y^2
$=(x^2+y^2)^2+x^4+x^2y^2+y^2$
$=1+x^2(x^2+y^2)+y^2=1+x^2+y^2=1+1=2$