Lời giải:

$A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+1998$

$2A=2x^2+4y^2+6z^2-4xy+4xz-4x-4y-16z+3996$

$=(x^2-4xy+4y^2)+(x^2+4xz+4z^2)+2z^2-4x-4y-16z+3996$

$=(x-2y)^2+(x+2z)^2-4x-4y-16z+2z^2+3996$

$=(x-2y)^2+2(x-2y)+1+(x+2z)^2-6(x+2z)+9+2z^2-4z+3986$

$=(x-2y+1)^2+(x+2z-3)^2+2(z^2-2z+1)+3984$

$=(x-2y+1)^2+(x+2z-3)^2+2(z-1)^2+3984\geq 3984$

$\Rightarrow A\geq 1992$

Vậy $A_{\min}=1992$
Giá trị này đạt tại $x-2y+1=x+2z-3=z-1=0$

$\Leftrightarrow x=y=z=1$

Lời giải:

$A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+1998$

$2A=2x^2+4y^2+6z^2-4xy+4xz-4x-4y-16z+3996$

$=(x^2-4xy+4y^2)+(x^2+4xz+4z^2)+2z^2-4x-4y-16z+3996$

$=(x-2y)^2+(x+2z)^2-4x-4y-16z+2z^2+3996$

$=(x-2y)^2+2(x-2y)+1+(x+2z)^2-6(x+2z)+9+2z^2-4z+3986$

$=(x-2y+1)^2+(x+2z-3)^2+2(z^2-2z+1)+3984$

$=(x-2y+1)^2+(x+2z-3)^2+2(z-1)^2+3984\geq 3984$

$\Rightarrow A\geq 1992$

Vậy $A_{\min}=1992$
Giá trị này đạt tại $x-2y+1=x+2z-3=z-1=0$

$\Leftrightarrow x=y=z=1$

\(x^4+x^2+1=\left(x^2\right)^2+2.x^2.1+1^2=\left(x^2+1\right)^2\)

\(#NqHahh\)

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

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

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

\(x^3-2x^2+5x-4\)

\(=x^3-x^2-x^2+x+4x-4\)

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

Bạn xem lại đề, là $13x^3$ hay $13x^2$ vậy?

Lời giải:
Trước hết ta cần nắm 1 số tính chất:

- Một scp lẻ khi chia 8 dư 1 (bạn có thể xét mô đun 4 của số đó để chứng minh)

- Một scp khi chia 5 dư $0,1$ hoặc $4$.

----------------------

Ta có: $2n+7$ là scp lẻ nên $2n+7\equiv 1\pmod 8$

$\Rightarrow 2n+6\equiv 0\pmod 8$

$\Rightarrow n+3\equiv 0\pmod 4$

$\Rightarrow n$ lẻ. 

$\Rightarrow 3n+10$ cũng là scp lẻ.

$\Rightarrow 3n+10\equiv 1\pmod 8$

$\Rightarrow 3n+9\equiv 0\pmod 8$

$\Rightarrow 3(n+3)\equiv 0\pmod 8\Rightarrow n+3\equiv 0\pmod 8(*)$

Lại có:

Đặt $2n+7=a^2, 3n+10=b^2$ với $a,b$ là số tự nhiên.

$\Rightarrow 2n+7+3n+10=a^2+b^2$

$\Rightarrow a^2+b^2=5n+17\equiv 2\pmod 5$

Ta thấy $a^2\equiv 0,1,4\pmod 5; b^2\equiv 0,1,4\pmod 5$

Do đó để $a^2+b^2\equiv 2\pmod 5$ thì chỉ khi $a^2, b^2\equiv 1\pmod 5$

$\Rightarrow 3n+10\equiv 1\pmod 5$

$\Rightarrow 3n+9\equiv 0\pmod 5$

$\Rightarrow 3(n+3)\equiv 0\pmod 5$

$\Rightarrow n+3\equiv 0\pmod 5(**)$

Từ $(*); (**)$ mà $(5,8)=1$ nên $n+3\vdots 40$.

a: Xét ΔABC có D,E lần lượt là trung điểm của AC,AB

=>DE là đường trung bình của ΔABC

=>DE//BC và \(DE=\dfrac{BC}{2}=2\left(cm\right)\)

Xét hình thang BEDC có

M,N lần lượt là trung điểm của EB,DC

=>MN là đường trung bình của hình thang BEDC

=>MN//ED//BC và \(MN=\dfrac{ED+BC}{2}=\dfrac{2+4}{2}=3\left(cm\right)\)

b: Xét ΔBED có MP//ED
nên \(\dfrac{MP}{ED}=\dfrac{BM}{BE}=\dfrac{1}{2}\)

=>\(MP=\dfrac{1}{2}ED=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot BC=\dfrac{1}{4}BC\)

Xét ΔCED có NQ//ED
nên \(\dfrac{NQ}{ED}=\dfrac{CN}{CD}=\dfrac{1}{2}\)

=>\(NQ=\dfrac{1}{2}ED=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot BC=\dfrac{1}{4}BC\)

\(MN=\dfrac{1}{2}\left(ED+BC\right)=\dfrac{1}{2}\left(\dfrac{1}{2}BC+BC\right)=\dfrac{1}{2}\cdot\dfrac{3}{2}BC=\dfrac{3}{4}BC\)

=>\(MP+PQ+QN=\dfrac{3}{4}BC\)

=>\(PQ=\dfrac{3}{4}BC-\dfrac{1}{4}BC-\dfrac{1}{4}BC=\dfrac{1}{4}BC\)

Do đó:MP=PQ=QN

13: \(x^3-6x^2+12x-8\)

\(=x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3\)

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

14: \(8x^3+12x^2y+6xy^2+y^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

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

16: \(x^{10}-1=\left(x^5-1\right)\left(x^5+1\right)=\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)\)

17: \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)

18: \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

19: \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)

20: \(9x^2+6xy+y^2=\left(3x\right)^2+2\cdot3x\cdot y+y^2=\left(3x+y\right)^2\)

21: \(6x-9-x^2=-\left(x^2-6x+9\right)\)

\(=-\left(x^2-2\cdot x\cdot3+3^2\right)\)

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

22: \(x^2+4xy+4y^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)

23: \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=2x\cdot2y=4xy\)

24: \(\left(x+y+z\right)^2-4z^2\)

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

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

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

25: \(\left(3x+1\right)^2-\left(x+1\right)^2\)

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

\(=2x\left(4x+2\right)=4x\left(2x+1\right)\)

26: \(x^3y^3+125=\left(xy\right)^3+5^3\)

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

27: \(8x^3-y^3-6xy\left(2x-y\right)\)

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

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

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

28: \(\left(3x+2\right)^2-2\left(x-1\right)\left(3x+2\right)+\left(x-1\right)^2\)

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

\(=\left(2x+3\right)^2\)

a: Ta có: \(\widehat{ICA}+\widehat{ICB}=\widehat{ACB}=90^0\)

\(\widehat{ICB}+\widehat{NCB}=\widehat{NCI}=90^0\)

Do đó: \(\widehat{ICA}=\widehat{NCB}\)

Ta có: \(\widehat{CAI}+\widehat{CBI}=90^0\)(ΔCBA vuông tại C)

\(\widehat{CBI}+\widehat{CBN}=\widehat{NBI}=90^0\)

Do đó: \(\widehat{CAI}=\widehat{CBN}\)

Xét ΔCAI và ΔCBN có

\(\widehat{CAI}=\widehat{CBN}\)

\(\widehat{ICA}=\widehat{NCB}\)

Do đó: ΔCAI~ΔCBN

b: Ta có: \(\widehat{ACM}+\widehat{ACI}=\widehat{ICM}=90^0\)

\(\widehat{ICA}+\widehat{ICB}=\widehat{ACB}=90^0\)

Do đó: \(\widehat{ACM}=\widehat{ICB}\)

Ta có: \(\widehat{CAM}+\widehat{CAB}=\widehat{BAM}=90^0\)

\(\widehat{CAB}+\widehat{CBA}=90^0\)(ΔCAB vuông tại C)

Do đó: \(\widehat{CAM}=\widehat{CBA}\)

Xét ΔCAM và ΔCBI có

\(\widehat{CAM}=\widehat{CBI}\)

\(\widehat{ACM}=\widehat{BCI}\)

Do đó: ΔCAM~ΔCBI

=>\(\dfrac{AC}{CB}=\dfrac{AM}{BI}\)

=>\(AC\cdot BI=MA\cdot BC\)

c: Xét tứ giác CIBN có \(\widehat{ICN}+\widehat{IBN}=90^0+90^0=180^0\)

nên CIBN là tứ giác nội tiếp

=>\(\widehat{CIN}=\widehat{CBN}\)

=>\(\widehat{CIN}=\widehat{BAC}\)

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

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

=>\(x^2+4x+4-x^2+x+2-3=0\)

=>5x+3=0

=>5x=-3

=>\(x=-\dfrac{3}{5}\)

b: \(\left(2x+3\right)^2-4\left(x-1\right)^2=0\)

=>\(\left(2x+3\right)^2-\left(2x-2\right)^2=0\)

=>\(\left(2x+3+2x-2\right)\left(2x+3-2x+2\right)=0\)

=>\(5\left(4x+1\right)=0\)

=>4x+1=0

=>4x=-1

=>\(x=-\dfrac{1}{4}\)

c: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2+2\right)-2=0\)

=>\(x^3+1-x^3-2x-2=0\)

=>-2x-1=0

=>-2x=1

=>\(x=\dfrac{1}{-2}=-\dfrac{1}{2}\)