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) = \(\dfrac{2}{9}\)

C = \(x^3\) + \(x^2\).y - 2\(x^2\) - \(xy\) - y\(^2\) + 3y + \(x\) - 1

C = (\(x^3\) + \(x^2\).y - 2\(x^2\)) - (\(xy\) + y\(^2\) - 2y) + (y + \(x\) - 2) + 1

C = \(x^2\).(\(x\) + y - 2) - y(\(x\) + y - 2) + (y + \(x\) - 2) + 1 (1)

Thay \(x+y-2\) vào biểu thức (1) ta có:

C = \(x^2\). 0 - y . 0 + 0 + 1

C = 0 - 0 + 0 + 1

C = 1

D = \(x\).(\(x^3\) - y)(\(x^3\) - 2y\(^2\))(\(x^3\) - 3y\(^2\))(\(x^3\) - 4y\(^4\)) (1)

Thay \(x\) = 2 và y = - 2 vào biểu thức (1) ta có:

D = 2.(2\(^3\)+2).(2\(^3\)- 2.(-2\()^2\)).(2\(^3\)-3(-2)\(^2\))(2\(^3\)- 4.(-2)\(^4\))

D = 2.(2\(^3\)+ 2).(8 - 8).(2\(^3\)- 3(-2)\(^2\))(2\(^3\)- 4.(-2)\(^4\))

D = 2.(2\(^3\)+ 2).0.(2\(^3\)- 3(-2)\(^2\))(2\(^3\)- 4.(-2)\(^4\))

D = 0

Vì \(\left(x-2\right)^4\ge0\forall x\)dấu "=" xảy ra \(\Leftrightarrow\)x-2=0 \(\Leftrightarrow\)x=2

\(\left(2y-1\right)^{2014}\ge0\forall y\)Dấu "=" xảy ra \(\Leftrightarrow\)2y - 1=0 \(\Leftrightarrow y=\frac{1}{2}\)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2014}\ge0\)

Kết hợp với điều kiện đề bài \(\left(x-1\right)^4+\left(2y-1\right)^{2014}\le0\), ta được:

\(\left(x-2\right)^4+\left(2y-1\right)^{2014}=0\)

Vậy x = 2; \(y=\frac{1}{2}\)

Thay x=2; \(y=\frac{1}{2}\)vào M, ta có:

\(M=21.2^2.\frac{1}{2}+4.2.\left(\frac{1}{2}\right)^2\)

\(=21.4.\frac{1}{2}+4.2.\frac{1}{4}\)

\(=42+2=44\)

Vậy M=44

B=(x+y)²-2(x+y)=9-6=3
 

\(B=x^2+2xy+y^2-2x-2y\)

\(=\left(x^2+2xy+y^2\right)-\left(2x+2y\right)\)

\(=\left(xx+xy+xy+yy\right)-2\left(x+y\right)\)

\(=\left[x\left(x+y\right)+y\left(x+y\right)\right]-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)^2-2\left(x+y\right)\)

\(=3^2-2.3=9-6=3\)

\(2a^2+2b^2=5ab\)

<=>   \(2a^2+2b^2-5ab=0\)

<=>  \(2a^2-4ab-ab+2b^2=0\)

<=>   \(2a\left(a-2b\right)-b\left(a-2b\right)=0\)

<=>  \(\orbr{\begin{cases}2a=b\\a=2b\end{cases}}\)

Do b > a > 0

=>  b = 2a

\(A=\frac{a+b}{a-b}=\frac{a+2a}{a-2a}=\frac{3a}{-a}=-3\)

\(2a^2+2b^2=5ab\)

<=>   \(2a^2+2b^2-5ab=0\)

<=>  \(2a^2-4ab-ab+2b^2=0\)

<=>   \(2a\left(a-2b\right)-b\left(a-2b\right)=0\)

<=>  \(\left(2a-b\right)\left(a-2b\right)=0\)

<=>  \(\orbr{\begin{cases}2a-b=0\left(L\right)\\a-2b=0\end{cases}}\)

=>  \(a=2b\)

=>  \(A=\frac{a+2b}{2a-b}=\frac{2b+2b}{2.2b-b}=\frac{4b}{3b}=\frac{4}{3}\)