đề bài bảo j vậy

\(2x^2+xy-10x-5y\)

\(=\left(2x^2+xy\right)-\left(10x+5y\right)\)

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

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

\(x^2-y^2+9-6x\)

\(=\left(x^2-6x+9\right)-y^2\)

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

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

x² + 2xy +7x+ 7y + y² + 10

=x²+2xy+y²+7.(x+y)+10

=(x+y)²+7.(x+y)+10

=(x+y)²+2(x+y)+5(x+y)+10

=(x+y)(x+y+2)+5.(x+y+2)

=(x+y+2)(x+y+5)

= (x+y)(x+y) + 7(x+y) + 10

= (x+y+7)(x+y) + 10 

chắc là sai :)

chờ tí vừa làm xong 1 bài dài lòi mắt

1, B

2, A

3, A

4, C

5, A

6, B

7, D

8, D

9, D

10, A

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Bạn tham khảo với link trên nha!~

#Học tốt#

A=\(x^2+2y^2+7x+7y+12=x^2+2xy+y^2+7\left(x+y\right)+12+y^2\)

   \(=\left(x+y\right)^2+2\dfrac{7}{2}\left(x+y\right)+\left(\dfrac{7}{2}\right)^2-\dfrac{1}{4}+y^2\)

    \(=\left(x+y+\dfrac{7}{2}\right)^2+y^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)

Vậy Min A =\(\dfrac{-1}{4}\) .Dấu = xảy ra\(\left\{{}\begin{matrix}x+y+\dfrac{7}{2}=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\y=0\end{matrix}\right.\)

a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

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

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

c) Đổi đề: \(a^2x+a^2y-7x-7y\)

\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)

d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)

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

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

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

h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)

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

a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

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

d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)

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

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

h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

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