ta có A=x²-2xy+y²=x²+2xy+y²-4xy=(x+y)²-4 x (-10)=3²+40=49

          B=x²+y²=>B=x²+2xy+y²-2xy=(x+y)²-2 x (-10)=9+20=29

Bài 2:

\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

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

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

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

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)

a ) \(x^2-2xy+y^2-1\)

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

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

\(=9-1\)

\(=8\)

b ) \(x^2+y^2\)

\(=x^2-20+y^2+20\)

\(=x^2-2.10+y^2+20\)

\(=x^2-2xy+y^2+20\)

\(=\left(x-y\right)^2+20\)

\(=\left(-3\right)^2+20\)

\(=29\)

a) \(x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

b) Có: \(x^2-2xy+y^2=9\)

=> \(x^2+y^2=9+2xy=9+2\cdot10=9+20=29\)

help me plss

Bài 2: 

\(\dfrac{x^4-x^3+3x^2-x+a}{x^2-x+2}\)

\(=\dfrac{x^4-x^3+2x^2+x^2-x+2+a-2}{x^2-x+2}\)

\(=x^2+1+\dfrac{a-2}{x^2-x+2}\)

Để A chia hết cho B thì a-2=0

hay a=2

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

b: \(=-xy^2\cdot x^2-xy^2\cdot z=-xy^2\left(x^2+z\right)\)

c: x^2y^3-xy^2

=xy^2*xy-xy^2

=xy^2(xy-1)

d: -x^3z-z

=z(-x^3-1)

=-z(x+1)(x^2-x+1)

e: =x(x-y)+(x-y)

=(x-y)(x+1)

n: =x^2(x-1)-(x-1)

=(x-1)(x^2-1)

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

a: \(\dfrac{xy}{x^2+y^2}=\dfrac{5}{8}\)

=>\(\dfrac{xy}{5}=\dfrac{x^2+y^2}{8}=k\)

=>\(xy=5k;x^2+y^2=8k\)

\(A=\dfrac{8k-2\cdot5k}{8k+2\cdot5k}=\dfrac{-2}{18}=\dfrac{-1}{9}\)

b: Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\)

=>x=a*k; y=b*k; z=c*k

\(B=\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2}\)

\(=\dfrac{k^2\cdot\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

kinh nhờ học nhà thầy Khánh à ?

mấy bạn biết thầy Khánh ak thầy mk đó

a)

Sửa đề: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

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

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

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

Thay x-y=7 vào biểu thức (1), ta được:

\(A=\left(7+1\right)^2+36=8^2+36=100\)

Vậy: 100 là giá trị của biểu thức \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\) tại x-7=7

a, x² - 2xy + y² - 1

= (x - y)² - 1

= (-3)² - 1

= 9 - 1 

= 8

b, x² + y²

= x² - 20 + y² + 20

= x² - 2.10 + y² + 20

= x² - 2xy + y² + 20

= (x - y)² + 20

= (-3)² + 20

= 9 + 20

= 29