c: =>(2x+3y-1)^2+(2x-3y)=0

=>2x-3y=0 và 2x+3y=1

=>x=1/4; y=1/6

d: =>2y-3=0 và 2x+3y-1=0

=>y=3/2 và 2x=1-3y=1-9/2=-7/2

=>x=-7/4 và y=3/2

a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng

\(a,=3x\left(y-z\right)-y\left(y-z\right)=\left(3x-y\right)\left(y-z\right)\\ b,=x^3\left(x-1\right)+x\left(x-1\right)=x\left(x^2+1\right)\left(x-1\right)\\ c,=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\\ d,=\left(x-3\right)^2\\ e,=\left(x+2\right)^3\\ f,=\left(2x-x+y\right)\left(2x+x-y\right)=\left(x+y\right)\left(3x-y\right)\\ g,=\left(y+1\right)\left(5x-2\right)\\ h,=\left(x+2\right)^2\\ i,=x^2\left(x^2-2\right)\\ k,=3x\left(x-4y\right)\)

\(a,=\left(x+y\right)\left(y+z\right)\\ b,=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ c,=\left(x-y\right)\left(x+y\right)+\left(x-y\right)=\left(x+y+1\right)\left(x-y\right)\\ d,= \left(2x-5\right)\left(2x+5\right)\\ e,=\left(4y-3\right)\left(4y+3\right)\)

\(6x^2+18y^2+12x-12xy+9=0\)

\(\Rightarrow\left(2x^2-12xy+18y^2\right)+\left(4x^2+12x+9\right)=0\)

\(\Rightarrow2\left(x^2-6xy+9y^2\right)+\left(2x+3\right)^2=0\)

\(\Rightarrow2\left(x-3y\right)^2+\left(2x+3\right)^2=0\)

Vì \(\left\{{}\begin{matrix}2\left(x-3y\right)^2\ge0\\\left(2x+3\right)^2\ge0\end{matrix}\right.\) với mọi x,y

=> \(2\left(x-3y\right)^2+\left(2x+3\right)^2\ge0\)

Mà \(2\left(x-3y\right)^2+\left(2x+3\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}2\left(x-3y\right)^2=0\\\left(2x+3\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-3y=0\\2x+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3y=x\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=\dfrac{x}{3}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)

c) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-25\right]\)

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

d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

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

\(=\left(x-y\right)^2-\left(2m-n\right)^2\)

\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)

a) (4x²-3x-18)²-(4x²+3x)²

=(4x²-3x-18-4x²-3x)(4x²-3x-18+4x²+3x)

=(-6x-18)(8x²-18)

=-48x³+108x-144x²+324

b)B=27y^3-27y^2x+9yx^2-x^3 
= 27 . (1/3x)^3 - 27.(1/3x)².x + 9.1/3.x.x^2 - x^3 
= x^3 - 3x^3 + 3x^3 - x^3 
= 0

d) D=50y^2+x(x-2y)+14y(x-y) 

=50y^2 +x^2 -2xy +14xy -14y^2 

=36y^2 +x^2 +12xy 

=(6y + x)^2 

=81