\(x;y;z\rightarrow q;h;p\)

\(=\left(q^2+h^2+p^2\right)\left(q^2+h^2+p^2+2qh+2hp+2qp\right)+\left(qh+hp+pq\right)^2\)

\(Dat:\hept{\begin{cases}q^2+h^2+p^2=f\\qh+hp+qp=g\end{cases}}\Rightarrow\left(p^2+h^2+q^2\right)\left(p+q+h\right)^2+\left(qh+pq+ph\right)^2\)

\(=f\left(f+2g\right)+g^2=f^2+2fg+g^2=\left(f+g\right)^2=\left(q^2+h^2+p^2+qh+hp+pq\right)^2\)

shitbo Cho đệ sửa lại bài SP chứ bài SP dài quá ạ:p

\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)

\(=\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2+2xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)

Đặt \(x^2+y^2+z^2=a;xy+yz+zx=b\)

\(\Rightarrow a\left(a+2b\right)+b^2=a^2+2ab+b^2=\left(a+b\right)^2=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)

Làm cho mk đi @Ender Dragon Boy Vcl

(x−y+z)2+(z−y)2+2(x−y+z)(y−z)(x−y+z)2+(z−y)2+2(x−y+z)(y−z)

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

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

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

=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]

=(x−y+z)x+(z−y)(z−y−x+y−z)=(x−y+z)x+(z−y)(z−y−x+y−z)

=x2−xy+xz+(z−y)(−x)=x2−xy+xz+(z−y)(−x)

=x2−xy+xz−xz+xy=x2−xy+xz−xz+xy

=x2

Tu ke \(AH\perp BC\) Dat BH la x >0 

thi Xet tam giac AHB vuong tai H co

AH=\(\sqrt{2-x^2}\) cm   (DL PYTAGO)

=> CH = \(1+\sqrt{3}-x\) cm

Xet tam giac AHC vuong tai H co

\(AC^2=AH^2+HC^2\) Dinh Ly Pytago

<=> \(4=2-x^2+\left(1+\sqrt{3}-x\right)^2\)  

<=> \(4=2-x^2+1+3+x^2+2\sqrt{3}-2x-2\sqrt{3}x\)

<=> \(2\sqrt{3}-2\sqrt{3}x-2x+2=0\) 

<=> \(2\sqrt{3}\left(1-x\right)-2\left(1-x\right)=0\)

<=>\(\left(2\sqrt{3}-1\right)\left(1-x\right)=0\)

<=> x=1 

Suy ra \(AH=\sqrt{2-1}=1\)

cos B =\(\frac{BH}{AB}=\frac{1}{\sqrt{2}}\) => \(\widehat{B}=45^o\)

cos C=\(\frac{HC}{AC}=\frac{1+\sqrt{3}-1}{2}=\frac{\sqrt{3}}{2}=>\widehat{C}=30^o\)

Suy ra \(\widehat{A}=180^o-45^o-30^0=105^0\)

Study well

Cô-si ngược dấu thôi~~

Ta có:\(\sqrt{12a+\left(b-c\right)^2}=\frac{1}{\sqrt{12}}\cdot\sqrt{12\left[12a+\left(b-c\right)^2\right]}\)

\(\le\frac{1}{\sqrt{12}}\cdot\frac{12+12a+\left(b-c\right)^2}{2}\)

Tương tự ta có:
\(K\le\frac{1}{\sqrt{12}}\left(\frac{12+12a+\left(b-c\right)^2}{2}+\frac{12+12b+\left(a-c\right)^2}{2}+\frac{12+12c+\left(a-b\right)^2}{2}\right)\)

\(=\frac{1}{\sqrt{12}}\cdot\frac{36+12\left(a+b+c\right)+2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}{2}\)

Ta có:\(a^2+b^2+c^2\ge ab+bc+ca\) ( tự cm )

\(\Rightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)

\(\Rightarrow K\le\frac{1}{\sqrt{12}}\cdot36=6\sqrt{3}\)

P/S:Em ko chắc đâu ạ.sợ bị ngược dấu lắm.Nhất là đoạn cuối:((( 

\(\sqrt{12a+\left(b-c\right)^2}\le\sqrt{12a+\left(b+c\right)^2}=\sqrt{12a+\left(3-a\right)^2}=a+3\)

:) 

ko nghĩ đây là bài lớp 9

\(a+b=c\)

\(\Rightarrow\hept{\begin{cases}a=c-b\\b=c-a\\c=a+b\end{cases}}\)

Bằng các STN hửm ?

#Mật

\(a,x^8+14x^4+1=\left(x^8+14x^4+49\right)-48\)

\(=\left(x^4+7\right)^2-48\)

\(=\left(x^4+7+\sqrt{48}\right)\left(x^4+7-\sqrt{48}\right)\)

\(b,x^8+98x^4+1\)

\(=\left(x^8+98x^4+2401\right)-2400\)

\(=\left(x^4+49\right)^2-2400\)

\(=\left(x^4+49+20\sqrt{6}\right)\left(x^4+49-20\sqrt{6}\right)\)

Mình nghĩ vậy k bt đúng k :)

a) = a^10 - a + a^5 - a^2 + a^2 + a + 1

= a(a^9 - 1) + a^2(a^3 - 1) + (a^2 + a + 1)

= a.(a^3-1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)

= a.(a-1)(a^2 + a + 1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)

= (a^2 + a + 1)[(a.(a-1)(a^6 + a^3 + 1) + a^2 + 1]

b) x^5 - x^4 - 1 = x^5 - x^4 + x^3 - x^3 + x^2 - x - x^2 + x - 1

= x^3(x^2 - x + 1) - x(x^2 - x + 1) - (x^2 - x + 1)

= (x^2 - x + 1)(x^3 - x - 1)

a) \(a^{10}+a^5+1\)

\(=\left(a^{10}-a^9+a^7-a^6+a^5-a^3+a^2\right)\)

\(+\left(a^9-a^8+a^6-a^5+a^4-a^2+a\right)\)

\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(=a^2\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(+a\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

@hieu nguyen Em có nhân chéo hai vế và khai triển ra nhưng cũng không ra cái gì ạ. 

/nfvg6yguntbygynugytf6ht