x3+2x2y+xy2−9xx3+2x2y+xy2−9x

=x(x2+2xy+y2−9)=x(x2+2xy+y2−9)

=x[(x+y)2−32]=x(x+y+3)(x+y−3)=x[(x+y)2−32]=x(x+y+3)(x+y−3)

b) 2x−2y−x2+2xy−y22x−2y−x2+2xy−y2

=2(x−y)−(x−y)2=(x−y)(2−x+y)=2(x−y)−(x−y)2=(x−y)(2−x+y)

c) x2−2x−4y2−4yx2−2x−4y2−4y

=(x2−4y2)−(2x+4y)=(x2−4y2)−(2x+4y)

=(x−2y)(x+2y)−2(x+2y)=(x−2y)(x+2y)−2(x+2y)

=(x+2y)(x−2y−2)

mik gửi lời mời kết bạn cho cậu rồi đó

M=5x^2+y^2+2x(y-2)+2021

<=> M = 5x^2 + y^2 + 2xy - 4x + 2021

<=> M = 4x^2 + x^2 + y^2 + 2xy - 4x + 1 + 2020

<=> M = ( 4x^2 - 4x + 1) + (x^2 +2xy+y^2) +2020

<=> M =(2x-1)^2 + (x+y)^2 + 2020

Vì: (2x-1)^2 + (x+y)^2  > 0

=> (2x-1)^2 + (x+y)^2 + 2020 > 2020

Dấu "=" xảy ra khi và chỉ khi : (2x-1)^2=0 => 2x-1 = 0 => 2x= 1 => x=1/2

                                                (x+y)^2=0 => x+y=0 => 1/2 + y =0 => y= 0- 1/2 => y=-1/2

Vậy Mmin = 2020 khi x=1/2;y=-1/2

4x² + y² = 2
\(\hept{\begin{cases}4x+y=2\\4x-y=2\end{cases}}\)

\(\hept{\begin{cases}y=2-4x\\4x-\left(2-4x\right)=2\end{cases}}\)

\(\hept{\begin{cases}y=2-4x\\8x-2=2\end{cases}}\)

\(\hept{\begin{cases}y=0\\x=\frac{1}{2}\end{cases}}\)

Thay vào pt A ta có:

A=4.1/2 - 2.0

A=2-0

A=0

ko biết đk chứ GTLN GTNN là k biết r haha

Vì a + b + c = 0

<=> (a + b + c)² = 0

<=> a² + b² + c² = -2(ab + bc + ca)

Khi đó \(\frac{9\left(a^2+b^2+c^2\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}=\frac{-18\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}\)

\(=\frac{-18\left(ab+bc+ca\right)}{-6\left(ab+bc+ca\right)}=3\)

a² - 6b² = ab

<=> (a + 2b)(a - 3b) = 0

<=> \(\orbr{\begin{cases}a=-2b\left(\text{loại}\right)\\a=3b\left(tm\right)\end{cases}}\)

Khi đó \(A=\frac{2ab}{a^2-7b^2}=\frac{6b^2}{2b^2}=3\)