vì x=5+y => x-y=5

đặt \(A=x^2+y\left(y-2x\right)+75\)

          \(=x^2+y^2-2xy+75\)

             \(=\left(x-y\right)^2+75\)

            \(=5^2+75\)

               =100

b)   đặt \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+65\) 

                 \(=x^2+2x+y^2-2y-2xy+65\)

                   \(=\left(x^2+y^2-2xy\right)+\left(2x-2y\right)+65\)

                   \(=\left(x-y\right)^2+2\left(x-y\right)+65\)

                    \(=5^2+2.5+65\)

                     =100

Lần sau bạn ghi đúng lớp ạ

x( x + 2 ) + y( y - 2 ) - 2xy + 65

= x² + 2x + y² - 2y - 2xy + 65

= ( x² - 2xy + y² ) + ( 2x - 2y ) + 65

= ( x - y )² + 2( x - y ) + 65

Rồi x - y bằng bao nhiêu bạn thế vô

Good luck

           Tiếng Anh lớp 1  ?          

yyyyy

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

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

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

Ta có :  \(N=\left(x-y\right)^3-x^2+2xy-y^2\)

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

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

Thay  \(x-y=-5\)vào ta được :

\(N=\left(-5\right)^3-\left(-5\right)^2\)

\(N=-125-25\)

\(N=-150\)

Vậy  \(N=-150\)với  \(x-y=-5\)

a) Ta có: A = (x + y)³ + 2x² + 4xy + 2y²

    A = 7³ + 2(x² + 2xy + y²)

   A = 343 + 2(x + y)²

 A = 343 + 2. 7²

  A = 343 + 98 = 441

b) B = (x - y)³ - x² + 2xy - y²

=> B = (-5)³ - (x² - 2xy + y²)

=> B = -125 - (x - y)²

=> B = -125 - (-5)²

=> B = -125 - 25 = -150

\(P=2x^2y+2xy^2-2xy+5\)

\(P=2xy\left(x+y-1\right)+5\)

Thay x + y = 1 ta có :

\(P=2xy\left(1-1\right)+5\)

\(P=2xy\cdot0+5\)

\(P=5\)

Vậy....

\(A=\left(x+y\right)^2-8\left(x+y\right)+2\)
Thay x + y = 5 vào A, ta được:
\(A=5^2-8.5+2=-13\)
Vậy x + y = 5 thì A = -13

\(A=\left(x+y\right)^2-8\left(x+y\right)+16-14\\ A=\left(x+y-4\right)^2-14\\ A=\left(5-4\right)^2-14=-13\)

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

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

\(=5^2-8.5+2\)

\(=-13\)

Vậy \(A=-13\) khi x + y = 5

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

\(=7^3+2\left(x^2+2xy+y^2\right)\)

\(=343+2\left(x+y\right)^2\)

\(=343+2.7^2\)

\(=343+98=441\)

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

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

\(=-125-\left(-5\right)^2\)

\(=-125-25=-150\)

\(a,M=\left(x+y\right)^3+2x^2+4xy+2y^2\)

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

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

\(=\left(x+y\right)^2\left(x+y+2\right)=7^2.9=49.9=441\)

\(b,N=\left(x-y\right)^3-x^2+2xy-y^2\)

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

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

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

\(=\left(-5\right)^2\left(-5-1\right)=15.-6=-150\)