a, \(x^2+y^2=8\Rightarrow\left(x+y\right)^2-2xy=8\Rightarrow xy=\frac{8-\left(x+y\right)^2}{-2}=\frac{8-4}{-2}=-2\)

=>\(M=x^3+x^4+y^3+y^4=\left(x+y\right)^3-3xy\left(x+y\right)+\left(x^2+y^2\right)^2-2x^2y^2\)

\(=2^3-3.\left(-2\right).2+8^2-2.\left(-2\right)^2=76\)

b, \(M=x^2+y^2+2xy-4x-4y+3=\left(x+y\right)^2-4\left(x+y\right)+4-1=\left(x+y-2\right)^2-1=\left(5-2\right)^2-1=8\)

a)

Ta có:

\(2xy=(x+y)^2-(x^2+y^2)=2^2-8=-4\Rightarrow xy=-2\)

Vậy:

\(M=x^3+x^4+y^3+y^4=(x^3+y^3)+(x^4+y^4)\)

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

\(=2.8-(-2).2+8^2-2(-2)^2\)

\(=76\)

b)

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

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

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

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

\(=5.5-4.5+3=8\)

\(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{4}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)

\(=-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{18}=\dfrac{-9}{72}+\dfrac{12}{72}-\dfrac{4}{72}=-\dfrac{1}{72}\)

Câu b đề sai rồi bạn

a) \(3x^2-3xy-5x+5y\)

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

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

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

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)

1. Ta có: 

\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)

Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)

2.

a) \(x^2+4y^2+6x-12y+18=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(2y-3\right)^2=0\)

Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)

Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)

b) \(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)

.....................................

Rồi giải tương tự như trên

Ta có A=x(x+2)+y(y-2)-2xy+37 

            = x^2 +2x + y^2 - 2y - 2xy +37

            =(x^2 +y^2 -2xy +1 +2x - 2y) +36

            =(x -y +1)^2 +36

            = (7+1)^2 +36 = 64 +36 =100

tks thằng hoàng nhá =))

số cần tìm là 100

\(x\left(x+2\right)+y\left(y-2\right)-2xy+64=x^2+2x+y^2-2y-2xy+64\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+64\)\(=\left(x-y\right)^2+2\left(x-y\right)+64\)

\(=\left(x-y\right)\left(x-y+2\right)+64\)\(=\left(y+5-y\right)\left(y+5-y+2\right)+64\)

=5.7+64=99=99