1.A

2.C

3.B

4.C

a

c

b

c

ai ,mình tích  lại

2x^2+xy+2y^2 = 5/4.(x+y)^2 + 3/4. (x-y)^2 >= 5/4. (x+y)^2 
=> cbh(2x^2+xy+2y^2) >= cbh5 / 2. (x+y) 
tương tự với 2 căn còn lại.. cộng vế ta có VT >= cbh5 ( x+y+z) = cbh5 : dpcm 
dau = cay ra <=> x=y=z=1/3

\(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy.\left(x^2-y^2-2y-1\right)\)

\(=2xy.[x^2-\left(y^2+2y+1\right)]\)

\(=2xy.[x^2-\left(y+1\right)^2]\)

\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)

Vậy chọn đáp án A

chọn A

a. Ta có: x²+y²-2x+4y+5=0

⇌(x-1)²+(y-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

b. Ta có: 4x²+y²-4x-6y+10=0

⇌ (2x-1)²+(y-3)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)

c.Ta có: 5x²-4xy+y²-4x+4=0

⇌(2x-y)²+(x-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)

d.Ta có: 2x²-4xy+4y²-10x+25=0

⇌ (x-2y)²+(x-5)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)

a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)

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

b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)

c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)

d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)

\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)

e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)

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

f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)

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

1,

a, (2x + 1- x + 3)² = (x+4)²

b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)

c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)

d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)

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

f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)

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

Chúc các bn hc tốt

D ez nhất :v

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

\(=\left(x-1\right)^2+\left(y+2\right)^2+5\ge5\)

Đẳng thức xảy ra khi x = 1 và y = -2

\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)

\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)

\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)

Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1