\(a,1-y^3+6xy^2-12x^2y+8x^3=1-\left(y^3-6xy^2+12x^2y-8x^3\right)=1^3-\left(y-2x\right)^3=\left(1-y+2x\right)\left[1+1.\left(y-2x\right)+\left(y-2x\right)^2\right]=\left(1-y+2x\right)\left(1+y-2x+y^2-4x+4x^2\right)=\left(1-y+2x\right)\left(1+y-6x+y^2+4x^2\right)\)\(b,\left(x-z\right)^2-y^2+2y-1=\left(x-z\right)^2-\left(y^2-2x+1\right)=\left(x-z\right)^2-\left(y-1\right)^2=\left(x-z+y-1\right)\left(x-z-y+1\right)\)\(c,x^3+y^3+3y^2+3y+1=x^3+\left(y+1\right)^3=\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\)

\(\left(x+1\right)^2\left(x+2\right)+\left(x-1\right)^2\left(x-2\right)=12\)

\(\Leftrightarrow\left(x^2+2x+1\right)\left(x+2\right)+\left(x^2-2x+1\right)\left(x-2\right)=12\)\(\Leftrightarrow x^3+2x^2+2x^2+4x+x+2+x^3-2x^2-2x^2+4x+x-2-12=0\)\(\Leftrightarrow2x^3+10x-12=0\)

\(\Leftrightarrow2x^3-2x+12x-12=0\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+6\right)=0\)

Ta có :

\(x^2+x+6=x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\)\(\Rightarrow x-1=0\Leftrightarrow x=1\)

Vậy phương trình có tập nghiệm \(S=\left\{1\right\}\)

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

\(x^2+y^2+1\ge xy+x+y\)

\(\Leftrightarrow2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-2y+1+x^2-2x+1\ge0\)\(\Leftrightarrow\left(x-y\right)^2-\left(y-1\right)^2-\left(x-1\right)^2\ge0\)

Đúng với mọi x , y

Đẳng thức xảy ra khi \(\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\y=1\\x=1\end{matrix}\right.\Rightarrow x=y=1\)

1. His enthusiast for music has stayed strong, throughout his 23 years in radio.

2. He showed unswerving loyalty to his friends.

3. We all have great simpathy for the victims of the flood.

4intimate is very important because uncertain people cannot have a life long friendship.

5. It was very unselfish of him to offer us his room.

6. We were delighted by the wonderful hospitality of the local people.

7. The wine had made him incapable of thinking clearly.

8. Playing a game with the children is a good way of getting them acquainted

9. Inform the police immediately if you see anything suspicious

10. Friendships need time to develop.

\(C=b^3+c^3+ab^2+ac^2-abc=\left(b+c\right)\left(b^2-bc+c^2\right)+a\left(b^2-bc+c^2\right)=\left(b^2-bc+c^2\right)\left(a+b+c\right)\)Vì a + b + c = 0 \(\Rightarrow\left(a+b+c\right)\left(b^2-bc+c^2\right)=0\Rightarrow C=0\)

a)

Xét tam giác BMN và DPQ có:

\(BM=DP\)

\(\widehat{MDN}=\widehat{PDQ}\)

\(BN=DQ\)

\(\Rightarrow\) tam giác BMN = DPQ(c.g.c)
\(\Rightarrow MN=PQ\) (1)

Chứng minh tương tự, ta được: tam giác AQM = CNP\(\Rightarrow QM=NP\)(2)

Từ (1) và (2) \(\Rightarrow\) MNPQ là hình bình hành

-nếu a, b cùng chẵn thì ab(a+b) chẵn

-nếu a chẵn b lẻ thì ab(a+b) chẵn vì ab(a+b) chia hết cho 2

-nếu a lẻ b chẵn tương tự

-nếu a lẻ b lẻ thì a+b chãn => dpcm

vì ABCD là hình thang cân \(\Rightarrow\) AD = BD (1)

Vì tam giác ABD là tam giác cân tại A\(\Rightarrow\)AB = AD (2)

Vì tam giác BCD là tan giác cân \(\Rightarrow\) BC = AC (3)

Từ (1) ,(2), (3)\(\Rightarrow AB=BC=CD=AD\) \(\Rightarrow\) ABCD là hình vuông \(\Rightarrow\widehat{A}=\widehat{B}=\widehat{C}=\widehat{D}=90^o\)

\(Al\left(NO_3\right)x=213\Leftrightarrow27+14x+48x=213\Leftrightarrow62x=213-27=186\Rightarrow x=\dfrac{186}{62}=3\)