Ta có B = x³ + x²y - 2x² - xy² + 2xy + 2y + 2x - 2 - x²y

=> B = x³ + (x²y - x²y) - 2x² - xy² + 2xy + 2y + 2x - 2

=> B = x³ - 2x² - xy² + 2xy + 2y + 2x - 2

và x + y - 2 = 0 => x + y = 2 => x = 2 - y

Thế x = 2 - y vào biểu thức B, ta có:

x³ - 2x² - xy² + 2xy + 2y + 2x - 2 = (2 - y)³ - 2 (2 - y)² - (2 - y) y² + 2y (2 - y) + 2y + 2 (2 - y) -2

= (2 - y)² (2 - y) - 2 (2 - y)² + 2y² - y³ + 4y - 2y² + 2y + 4 - 2y - 2

= (2 - y)² (2 - y - 2) + (2y² - 2y²) - y³ + 4y + (2y - 2y) + (4 - 2)

= (2 - y)² (-y) - y³ + 4y + 2

Vậy giá trị biểu thức B là (2 - y)² (-y) - y³ + 4y + 2 khi x + y - 2 = 0.

\(\left|x-1\right|+\left(y+2\right)^{20}=0\)

=>x-1=0 và y+2=0

=>x=1 và y=-2

Thay x=1 và y=-2 vào X, ta được:

\(X=2\cdot1^5-5\cdot\left(-2\right)^3+2015\)

\(=2017+40=2057\)

\(A=x^2-2x-y+3y-1\)

\(B=-2x^2+3y^2-5x+y+3\)

a) \(A+B=\left(x^2-2x-y+3y-1\right)+\left(-2x^2+3y^2-5x+y+3\right)\)

\(=x^2-2x-y+3y-1-2x^2+3y^2-5x+y+3\)

\(=\left(x^2-2x^2\right)+3y^2+\left(-2x-5x\right)+\left(-y+3y+y\right)+3-1\)

\(=-x^2+3y^2-7x+3y+2\)

\(A-B=\left(x^2-2x-y+3y-1\right)-\left(-2x^2+3y^2-5x+y+3\right)\)

\(=x^2-2x-y+3y-1+2x^2-3y^2+5x-y-3\)

\(=\left(x^2+2x^2\right)-3y^2+\left(-2x+5x\right)+\left(-y+3y-y\right)-1-3\)

\(=3x^2-3y+3x+y-4\)

b) tại x=1 ; x=-2 ta có: 
\(A=1^2-2.1-\left(-2\right)+3.\left(-2\right)-1\)

\(A=1-2+2-6-1=-6\)

Vậy -6 là giá trị của đa thức A tại x=1 y=-2

a) \(A+B=\left(x^2-2x-y+3y-1\right)+\left(-2x^2+3y^2-5x+y+3\right)\)

                \(=-x^2+3y^2-7x+3y+2\)

\(A-B=\left(x^2-2x-y+3y-1\right)-\left(-2x^2+3y^2-5x+y+3\right)\)

           \(=3x^2-3y^2+3x+2y-4\)

b) \(A\left(1;-2\right)=1^2-2\cdot1-\left(-2\right)+3\cdot\left(-2\right)-1\)

                   \(=1-2+2-6-1\)

                   \(=-6\)

1) Tìm x và y biết

a) (2x+1)² + y² = 0

Ta có : \(\left(2x+1\right)^2\ge0;y^2\ge0\)

\(\Rightarrow\left(2x+1\right)^2+y^2\ge0\)

Để \(\left(2x+1\right)^2+y^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\y^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=0\end{matrix}\right.\)

b) x² + 2x + 1 + (y-1)² = 0

\(\Rightarrow\left(x+1\right)^2+\left(y-1\right)^2=0\)

Lập luận tương tự câu a ,ta có :

\(\left(x+1\right)^2+\left(y-1\right)^2\ge0\)

\(\left(x+1\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

c) x² - 2x + y² + 4y + 5 = 0

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

Lập luận tương tự 2 câu trên

\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

A =19^3+3.19^2+3.19+2

A =6859+1083+59

A =8001

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)