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

M=x^3+x^2.y-2x^2-xy-y^2+3y+x-1

=>  M=x^2﴾x+y‐2﴿‐﴾xy+y^2‐2y﴿+﴾y+x‐1﴿ = 0‐ y﴾x+y‐2﴿+1=1

N=x^3‐2x^2‐xy^2+2xy+2y+2x‐2

=> N= 2﴾x+y‐1﴿+x﴾x^2‐y^2﴿‐2x﴾x‐y﴿=2+x﴾x+y﴿﴾x‐y﴿‐2x﴾x‐y﴿=2+﴾x^2+xy‐2x﴿﴾x‐y﴿=2+x﴾x+y‐2﴿﴾x‐ y﴿=2+0=2﴾vì x+y‐2=0﴿ 

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

\(M=x^3+x^2y-xy^2-y^3+x^2-y^2+2x+2y+3\)

\(M=\left(x^3-y^3\right)+\left(x^2y-xy^2\right)+\left(x^2-y^2\right)+\left(2x+2y+2\right)+1\)

\(M=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)+\left(x-y\right)\left(x+y\right)+2\left(x+y+1\right)+1\)

\(M=\left(x-y\right)\left(x^2+xy+y^2+xy+x+y\right)+2.0+1\)

\(M=\left(x-y\right)\left[\left(x+y\right)^2+\left(x+y\right)\right]+1\)

\(M=\left(x-y\right)\left(x+y\right)\left(x+y+1\right)+1\)

\(M=\left(x-y\right)\left(x+y\right).0+1\)

\(M=1\)

Ở bài này mk áp dụng hằng đẳng thức (a³-b³)=(a-b)(a²+ab+b²) ,(a²-b²)=(a-b)(a+b);(a²+2ab+b²)=(a+b)²

MIK nghĩ bạn nên tra ông google nha 

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