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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) = \(\dfrac{2}{9}\)
a) x^2 -5x tại x=1,x=-1,x=1 phần 2
Thay x=1 =>\(1^2-5.1=1-5=-4\)
Thay \(x=-1\Rightarrow\left(-1\right)^2-5.\left(-1\right)=1+5=6\)
Thay \(x=\frac{1}{2}\Rightarrow\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)=\frac{1}{4}-\frac{5}{2}=-\frac{9}{4}\)
b)3x^2-xy tại x= -3,y=-5
Thay \(x=-3;y=-5\Rightarrow3.\left(-3\right)^2-\left(-3\right).\left(-5\right)=3.9-15=12\)
c)5-xy^3 tại x=1,y=-3
\(Thay...x=1;y=-3\Rightarrow5-1.\left(-3\right)^3=5-1.\left(-27\right)=5+27=32\)
d)x^5-5 tại x=1,-1
\(Thay..x=1\Rightarrow1^5-5=1-5=-4\)
\(Thay..x=-1\Rightarrow\left(-1\right)^5-5=-1-5=-6\)
e)x^2-3x-5 tại x=-2,y=-1
\(Thay.x=-2;y=-1\Rightarrow\left(-2\right)^2-3\left(-2\right)-5=5+6-5=6\)
g)x^2y^2+x^4y^4+x^6y^6 tại x=1,y=-1
\(Thayx=1;y=-1\Rightarrow1^2\left(-1\right)^2+1^4\left(-1\right)^4+1^6\left(-1\right)^6=1+1+1=3\)
Thay x=1,y=-1 vào A ta có A=-12.(-1)3 (-2.1.(-1)2 )2=4. Chọn D