\(a,VT=\dfrac{3y\cdot2x}{4\cdot2x}=\dfrac{6xy}{8x}=VP\\ b,VT=\dfrac{\left(x+y\right)\cdot3a\left(x+y\right)}{3a\cdot3a\left(x+y\right)}=\dfrac{3a\left(x+y\right)^2}{9a^2\left(x+y\right)}=VP\)

ta có: \(\dfrac{x^2+y^2}{xy}=\dfrac{2}{3}\Rightarrow2xy=3x^2+3y^2\\ \Rightarrow6xy=9x^2+9y^2\)

thay vào M, ta được:

\(M=\dfrac{x^2+9x^2+9y^2+y^2}{x^2-9x^2-9y^2+y^2}=\dfrac{10x^2+10y^2}{-8x^2-8y^2}\\ M=\dfrac{10\left(x^2+y^2\right)}{-8\left(x^2+y^2\right)}=\dfrac{10}{-8}=-\dfrac{5}{4}\)

Ta có : 6xy.( xy - y2 ) - 8x2 .( x - y2 ) + 5y2 ( x2 - xy )

=6x²y²-6xy³-8x³+8x²y²+5x²y²-5xy³

⁼ (6x²y²+8x²y²+5x²y²) +(-6xy³-5xy³)-8x³

=19x²y²-11xy³-8x³

Thay x= 1/2, y=2 ta đc :

19.1/2².2²-11.1/2.2³-8.1/2³

=19.1/4.4-11.1/2.8-8.1/8

=19-11.4-1

=-36

\(A=6xy\left(xy-y^2\right)-8x^2\left(x-y\right)^2+5y\left(x^2-xy\right)\)

\(=6xy^2\left(x-y\right)-8x^2\left(x-y\right)\left(x-y\right)+5xy\left(x-y\right)\)

\(=x\left(x-y\right)\left(6y^2-8x\left(x-y\right)+5y\right)\)

\(=x\left(x-y\right)\left(6y^2-8x^2+8xy+5y\right)\)

\(=x\left(x-y\right)\left[2\left(3y+2x\right)\left(y-2x\right)+16xy+5y\right]\)

Thay x=1/2; y =2 ta được

\(A=\frac{1}{2}\left(\frac{1}{2}-2\right)\left[0+16\cdot\frac{1}{2}\cdot2+5\cdot2\right]=-\frac{1}{2}\cdot\frac{3}{2}\cdot26=-\frac{39}{2}\).

6xy ( xy - y² ) - 8x² ( x - y )² + 5y ( x² - xy ) 

= 6x²y² - 6xy³ - 8x³ + 8x²y + 5yx² - 5xy²

= xy ( 6xy - 6y² + 8x + 5x - 5y ) - 8x³

Thay x= \(\frac{1}{2}\)     ; y = 2

= 6 - 6.4 + 8. \(\frac{1}{2}\)      + 5. \(\frac{1}{2}\)       -  5.2 - 8.8

=> 6 - 24 + 4 + 2,5 - 10 - 64

= - 85,5

\(6xy\left(xy-ỳ^2\right)-8x^2\left(x-y\right)^2+5y\left(x^2-xy\right)\)

\(=6x^2y^2-6xy^3-8x^2\cdot\left(x^2-y^2\right)+5x^2y-5xy^2\)

\(=6x^2y^2-6xy^3-8x^4+8x^2y^2+5x^2y-5xy^2\)

\(=-8x^4+14x^2y^2+5x^2y-6xy^3-5xy^2\)

 Thay \(x=\frac{1}{2};y=2\) vào đa thức \(-8x^4+14x^2y^2+5x^2y-6xy^3-5xy^2\)

\(-8\left(\frac{1}{2}\right)^4+14\left(\frac{1}{2}\right)^2\cdot2^2+5\left(\frac{1}{2}\right)^2\cdot2-6\cdot\frac{1}{2}\cdot2^3-5\cdot\frac{1}{2}\cdot2^2\)

\(=-\frac{1}{2}+14+\frac{5}{2}-24-10\)

\(=-18\)

\(A=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}=\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}+\dfrac{3\left(x^2+y^2\right)}{4xy}\)

\(A\ge2\sqrt{\dfrac{\left(x^2+y^2\right)xy}{4xy\left(x^2+y^2\right)}}+\dfrac{3.2xy}{4xy}=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)

\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)

\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(\frac{-6xy\left(x+y\right)^2}{8x^3y\left(x+y\right)}=\frac{-3\left(x+y\right)}{4x^2}\)

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3