a) Ta có:

\(A=x^2+2xy+y^2-4x-4y+1\)

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

Thay x + y = 3 vào A

\(A=3^2-4.3+1\)

\(A=9-12+1\)

\(A=-2\)

b) Sửa đề:

\(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(B=x^2+2x+y^2-2y-2xy+37\)

\(B=\left(x^2+y^2+1+2x-2y-2xy\right)+36\)

\(B=\left(x-y+1\right)^2+36\)

Thay x - y = 7 vào B

\(B=\left(7+1\right)^2+36\)

\(B=100\)

c) Ta có:

\(C=x^2+4y^2-2x+10+4xy-4y\)

\(C=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

\(C=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay x + 2y = 5 vào C

\(C=5^2-2.5+10\)

\(C=25-10+10\)

\(C=25\)

a, \(A=x^2+2xy+y^2-4x-4y+1\)

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

Thay x + y = 3

\(\Leftrightarrow A=9-12+1=-2\)

Vậy A = -2 khi x + y = 3

b, \(B=x^2+4y^2-2x+10+4xy-4y\)

\(=x^2+4xy+4y^2-2x-4y+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay x + 2y = 5 có:
\(B=25-10+10=25\)

Vậy B = 25 khi x + 2y = 5

Câu 9: D

Câu 10: A

Yêu cầu đề là gì vậy bạn?

a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=x^3+8y^3-x^3+y^3\)

\(=9y^3\)

b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)

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

\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)

\(=x^3-x^2-x+1-x^3-8\)

\(=-x^2-x-7\)

Câu 2: 

\(B=x^2+2x+y^2-2x-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2\cdot7+37=49+37+14=100\)

Câu 3: 

\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2\cdot5+10=25\)

\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a,9x^2+y^2+2z^2−18x+4z−6y+20=0

⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0

⇔x=1;y=3;z=−1

b,5x^2+5y^2+8xy+2y−2x+2=0

⇔4(x+y)2+(x−1)2+(y+1)2=0

⇔x=−y;x=1y=−1⇔x=1y=−1

c,5x^2+2y^2+4xy−2x+4y+5=0

⇔(2x+y)^2+(x−1)^2+(y+2)^2=0

⇔2x=−y;x=1;y=−2

⇔x=1;y=−2

⇔(x−1)^2+(2y−3)^2+(z+2)^2=0

\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương

a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)

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

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

b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)

c: \(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(y-x\right)+3=3\)

d: \(=\left(x+y\right)^2-4\left(x+y\right)+1\)

=9-12+1

=-2