\(A=5xy^2-3x^2y+6x+7y^2+1\)

\(B=13xy^2-6x^2y+3y^2+5x+5\)

=>\(A+B=18xy^2-9x^2y+11x+10y^2+6\)

\(A-B=-8xy^2+3x^2y+x+4y^2-4\)

a) \(2xy+2x-y=8\)

\(\Rightarrow\ 2x\left(y+1\right)-\left(y+1\right)=7\)

\(\Leftrightarrow\left(2x-1\right)\left(y+1\right)=7\)

\(\Rightarrow\left[\begin{matrix}\begin{cases}2x-1=-7\\y+1=-1\end{cases}\\\begin{cases}2x-1=-1\\y+1=-7\end{cases}\end{matrix}\right.\left[\begin{matrix}\begin{cases}2x-1=7\\y+1=1\end{cases}\\\begin{cases}2x-1=1\\y+1=7\end{cases}\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=4\\y=0\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=1\\y=6\end{cases}\\\left[\begin{matrix}\begin{cases}x=-3\\y=-2\end{cases}\\\begin{cases}x=0\\y=-8\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

c)\(x^2+xy+x+y=2\)

\(\Leftrightarrow x\left(x+1\right)+y\left(x+1\right)=2\)

\(\Leftrightarrow\left(x+y\right)\left(x+1\right)=2\)

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

d)\(2x^2+7xy+6y^2=15\)

\(\Leftrightarrow\left(2x+3y\right)\left(x+2y\right)=15\)

\(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\left[\begin{matrix}\begin{cases}2x+3y=1\\x+2y=15\end{cases}\\\begin{cases}2x+3y=15\\x+2y=1\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}2x+3y=3\\x+2y=5\end{cases}\\\begin{cases}2x+3y=5\\x+2y=3\end{cases}\end{matrix}\right.\end{matrix}\right.\\\left[\begin{matrix}\left[\begin{matrix}\begin{cases}2x+3y=-1\\x+2y=-15\end{cases}\\\begin{cases}2x+3y=-15\\x+2y=-1\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}2x+3y=-3\\x+2y=-5\end{cases}\\\begin{cases}2x+3y=-5\\x+2y=-3\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=-43\\y=29\end{cases}\\\begin{cases}x=27\\y=-13\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=-9\\y=7\end{cases}\\\begin{cases}x=1\\y=1\end{cases}\end{matrix}\right.\end{matrix}\right.\\\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=43\\y=-29\end{cases}\\\begin{cases}x=-27\\y=13\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=9\\y=-7\end{cases}\\\begin{cases}x=-1\\y=-1\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

Bài 1: 

a,  \(x^2\) +2\(x\) = 0

     \(x.\left(x+2\right)\) = 0

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

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

      \(x\) \(\in\) {-2; 0}

b, (-2.\(x\)).(-4\(x\)) + 28  = 100

      8\(x^2\)           + 28  = 100

        8\(x^2\)                   = 100 - 28

        8\(x^2\)                   = 72

          \(x^2\)                  = 72 : 8

          \(x^2\)                   = 9

           \(x^2\)                  = 3²

          |\(x\)|                  = 3

          \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\) 

Vậy \(\in\) {-3; 3}

c, 5.\(x\) (-\(x^2\)) + 1 = 6

   - 5.\(x^3\)       + 1 = 6

   5\(x^3\)                 = 1 - 6

   5\(x^3\)                 = - 5

    \(x^3\)                  =  -1

    \(x\)                    =  - 1

d, 3\(x^2\) + 12\(x\) = 0

   3\(x.\left(x+4\right)\) = 0

   \(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-4; 0}

e, 4.\(x.3\) = 4.\(x\)

    12\(x\) - 4\(x\) = 0

      8\(x\)          = 0

         \(x\)        = 0

a)\(\left(6x^2-3xy^2\right)+M=^2+y^2-2y^2\)

\(\Rightarrow M=\left(x^2+y^2-2xy^2\right)-\left(6x^2-3xy^2\right)\)

\(\Rightarrow M=x^2+y^2-2xy^2-6x^2+3xy^2\)

\(\Rightarrow M=\left(x^2-6x^2\right)+y^2+\left(-2xy^2+3xy^2\right)\)

\(\Rightarrow M=-7x^2+y^2+xy^2\)

b) \(M-\left(2xy-4y^2\right)=5xy+x^2-7y^2\)

\(\Rightarrow M=\left(5xy+x^2-7y^2\right)+\left(2xy-4y^2\right)\)

\(\Rightarrow M=5xy+x^2-7y^2+2xy-4y^2\)

\(\Rightarrow M=\left(5xy+2xy\right)+x^2+\left(-7y^2-4y^2\right)\)

\(\Rightarrow M=7xy+x^2-11y^2\)

Năng ceo à t lópw 7 r conf ko bt lm

phương trình nghiệm nguyên kiểu này liệt kê ước rồi kẻ bảng ra nhé

a) 

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

\(B=x^3+y^3=\left(x^3+3x^2y+3xy^2+y^3\right)-3x^2y-3xy^2=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3ab\)

\(C=x^5+y^5=\left(x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

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

\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-xy\left(x+y\right)\right)=a^5-5b\left(a^3-ab\right)\)

giup minh cau b o tren nha

Bài 2:

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

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

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

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

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

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

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

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

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

câu 1 a) xy=-5 => (x,y)=(1,-5),(-1,5)  

b) xy=-5 với x>y=>x=1,y=-5

c)(x+1)(y-2)=-5 => * x+1=1 và y-2=-5  => x=-1, y=-3

                              * x+1=-5 và y-2=1=> x=-6 , y=3

câu 2 , câu 3 tương tự

C1 : 

= x²y+xy²-x-y

=xy(x+y)-(x+y)

=(x+y)(xy-1)

c1 = xy(x+y) - (x+y)

= (x+y)(xy-1)

c2 đề sai

c3 = a(x²+y)-b(x²+y)

= (x²+y)(a-b)

mk làm mẫu cho mấy câu thui nha (mỏi tay quá) hiii. vì đây là toán dạng cơ bản 

c4:

x(x+1)²+x(x-5)-5(x+1)²

=(x+1)²(x-5)+x(x-5)

=(x-5)[(x+1)²+x)]

=(x-5)(x²+3x+1)

c5:

3x²-12y²

=3(x²-4y²)

=3(x-2y)(x+2y)

c6:

5xy²-10xyz+5xz²

=5x(y²-2yz+z²)

=5x(y-z)²