\(\frac{3x^2y-4x^2y}{2xy+5x}=\frac{-x^2y}{x\left(2y+5\right)}=\frac{-xy}{2y+5}\)

=3x²y-2x+5x

=x(3xy-2+5)

=x(3xy+3)

=3x(xy+1)

mk ko bt cs đúng hơn

\(A=4x^2-4xy+5y^2+20x-6y+2044\)

\(=\left(4x^2-4xy+y^2\right)+20x-6y+4y^2+2044\)

\(=\left(2x-y\right)^2+10\left(2x-y\right)+25+\left(4y^2+4y+1\right)+2018\)

\(=\left(2x-y+5\right)^2+\left(2y+1\right)^2+2018\ge2018\)

Dấu "=" xảy ra tại \(y=-\frac{1}{2};x=-\frac{11}{4}\)

Ta có \(A=4x^2-4xy+5y^2+20x-6y+2044\)

            \(=4x^2-4x\left(y-5\right)+\left(y-5\right)^2+4y^2+4y+1+2018\)

            \(=\left(2x-y+5\right)^2+\left(2y+1\right)^2+2018\)

Vì...\(\Rightarrow A\ge2018\)

Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y+5=0\\2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{11}{4}\\y=-\frac{1}{2}\end{cases}}}\)

lên mạng đấy nhé bạn

\(x^4+1\)

\(=x^3.x+1\)

\(=\left(x^3+1\right)\left(x+1\right)\)

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

\(x^4+1\)

\(=x^4+2x^2+1-2x^2\)

\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)

\(\Leftrightarrow\left(x^2+1-\sqrt{2}x\right)\left(x^2+1+\sqrt{2}x\right)\)

2019²-(2019-1)(2019+1)=2019²-(2019²-1)=1
k  :3

\(a,x^2-2005x-2006=0\)

\(\Leftrightarrow x^2+x-2006x-2006=0\)

\(\Leftrightarrow x\cdot\left(x+1\right)-2006\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2006\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2006=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2006\end{cases}}}\)

a) \(x^2-2005x-2006=0\)

Ta có: \(2005^2+4.2006=4028049\)

pt có 2 nghiệm:

\(x_1=\frac{2005+\sqrt{4028049}}{2}\);\(x_2=\frac{2005-\sqrt{4028049}}{2}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{2005+\sqrt{4028049}}{2};\frac{2005-\sqrt{4028049}}{2}\right\}\)

\(75^2+25^2-50.75\)

\(=75^2-2.25.75+25^2\)

\(=\left(75-25\right)^2\)

\(=50^2\)

\(=2500\)

\(12x^2y^3-10x^2y^3:5x^2y^2+4xy\left(1-3xy\right)^2\)

\(=12x^2y^3-2y+4xy\left(1-6xy+9x^2y^2\right)\)

\(=12x^2y^3-2y+4xy-24x^2y^2+36x^3y^3\)

\(x^3-y^3-36xy\)

\(=\left(x-y\right)^3+3xy\left(x-y\right)-36xy\)

\(=12^3+36xy-36xy\)

\(=1728\)