\(B=x\left(x+2\right)+y\left(y-2\right)-2xy+100\)

\(=x^2+2x+y^2-2y-2xy+100\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+100\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+99\)

\(=\left(x-y+1\right)^2+99\)

Vì \(x=y+6\Rightarrow x-y=6\Rightarrow x-y+1=7\)

Thay \(x-y+1=7\)vào \(B\), ta có :

\(B=7^2+99=49+99=148\)

\(KL:B=148\)tại \(x=y+6\)

\(x^2-2xy+y^2-5x+5y+6=\left(x-y\right)^2-5\left(x-y\right)+6=7^2-5.7+6=20\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

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

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

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

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

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

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

\(=5\left[5^2+3.\left(-6\right)\right]-25\)

\(=5\left[25-18\right]-25\)

\(=5.7-25=35-25=10\)

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

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

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

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

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

\(=\left(-5\right)\left[\left(-5\right)^2-6\right]-\left(-5\right)^2\)

\(=\left(-5\right)\left(25-6\right)-25\)

\(=\left(-5\right).21-25\)

\(=-105-25=-130\)

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

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

Đến đây thì ko bk lm nx

bài lớp mấy vậy mấy bạn

Ta có : \(x^2+2x+y^2-2y-2xy+65\)

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

Mà \(x=y+5\)

\(\Rightarrow x-y=5\)

- Thay x  - y = 5 vào đa thức trên ta được :

\(=\left(x-y\right)^2+2\left(x-y\right)+65=100\)

Vậy ...

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)

\(\dfrac{x}{x^2+2xy+y^2}+\dfrac{2y}{x+y}+\dfrac{y}{x^2+2xy+y^2}\)

\(=\dfrac{x+y}{\left(x+y\right)^2}+\dfrac{2y}{x+y}\)

\(=\dfrac{1}{x+y}+\dfrac{2y}{x+y}=\dfrac{2y+1}{x+y}\)