1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)

2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)

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

\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)

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

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

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

A. (Theo mình là -4xy thì mới rút gọn được)

B = (6x + y)^2

C = (2x)^2 = 4x^2

B

B

2)3x²-6xy+3y²=3(x²-2xy+y²)=3(x-y)²

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

5)(x+y)³-(x-y)³=[(x+y)-(x-y)][(x+y)²+(x+y)(x-y)+(x-y)²]=(x+y-x+y)(x²+2xy+y²+x²-y²+x²-2xy+y²)=2y(3x²+y²)

6)3x²-5x+2=3x²-2x-3x+2=(3x²-3x)-(2x-2)=3x(x-1)-2(x-1)=(x-1)(3x-2)

help mì 

Ta có:

\(15x^4y^4-M=10x^2y^4+6x^2y^4\)

\(\Leftrightarrow M=15x^4y^4-\left(10x^2y^4+6x^2y^2\right)\)

\(\Leftrightarrow M=15x^4y^4-16x^2y^4\)

Thay \(x=-\dfrac{1}{2};x=2\) vào M ta có:

\(M=15\cdot\left(-\dfrac{1}{2}\right)^4\cdot2^4-16\cdot\left(-\dfrac{1}{2}\right)^2\cdot2^4=-49\)

\(a,A=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\\ A=\left(5+7-36\right)\left(5+7+36\right)=-24\cdot48=-1152\\ b,B=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\\ B=\left(2+2\right)\left(2-2-1\right)=4\cdot\left(-1\right)=-4\)

5^2 là năm mũ hai ak

\(=3\cdot25-27:9+25\cdot4-18:9=75-3+100-2=72+100-2=170\)

=>\(\dfrac{3}{x-5}-\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{9-y\left(x-5\right)}{3\left(x-5\right)}=\dfrac{1}{6}\)

=>9-y(x-5)=1/2(x-5)

=>(x-5)(1/2+y)=9

=>(x-5)(2y+1)=18

=>\(\left(x-5;2y+1\right)\in\left\{\left(18;1\right);\left(-18;-1\right);\left(2;9\right);\left(-2;-9\right);\left(6;3\right);\left(-6;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(23;0\right);\left(-13;-1\right);\left(7;4\right);\left(3;-5\right);\left(11;1\right);\left(-1;-2\right)\right\}\)