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

      \(=\left(x+y\right)^2-4xy\)

        = 49

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

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

= 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: \(x^2-2xy+y^2=9\)

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

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

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)

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

Với x+y=3

=> A=\(3^2=9\)

b)

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

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

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

với x+y=3 ; xy=-10

=> B=\(3^2-4.\left(-10\right)\)

B=\(9+40=49\)

c)

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

   =\(\left(x+y\right)^2-2xy\)

   với x+y=3 ; xy=-10

=> C=\(3^2-2.\left(-10\right)\)

=>C=\(9+20=29\)

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

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

Thay \(x+y=3\)ta dc:

\(3^2=9\)

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

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

Thay \(x+y=3;xy=-10\)ta được:

\(3^2-4.\left(-10\right)=49\)

c,\(C=x^2+y^2+2xy-2xy\)

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

Thay \(x+y=3;xy=-10\)ta được

\(3^2-2.\left(-10\right)=29\)

Bài 8: Cho a+b= 1 nha ( mk thiếu đề)

Bài 1:

Theo bài ra ta có:

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

\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)

\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)

\(=25-10y+y^2+25-10x+x^2-4\)

\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)

\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)

\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)

\(=50-50+5^2-4-4\)

\(=25-8=17\)

Vậy giá trị của \(\left(x-y\right)^2\)là 17

ta có : M=2.(a^3  +b^3) -3.(a^2 + b^2)

       <=>M=2.(a+b)(a^2  -ab  +b^2)  - 3(a^2  +3b^2)

      <=>M=2(a^2  -ab  +b^2)  -3(a^2 +b^2)               vì a+b=1(gt)

      <=>M=-(a^2 +b^2 +2ab)

      <=>M=-(a+b)^2

      <=>M=-1  (vì a+b=1)