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1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
Ta có:
\(3\left(a^2+b^2+c^2\right)-3\left(a^2b+b^2c+c^2a\right)\)
= \(\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^2b+b^2c+c^2a\right)\)\(=a^3+ab^2+ac^2+a^2b+b^3+bc^2+ca^2+b^2c+c^3\)\(-3\left(a^2b+b^2c+c^2a\right)\)
\(=a^3+b^3+c^3+ab^2+bc^2+ca^2-2a^2b-2b^2c-2c^2a\)
\(=\left(a^3-2a^2b+ab^2\right)+\left(b^3-2b^2c+bc^2\right)+\left(c^3-2c^2a+ca^2\right)\)
\(=a\left(a-b\right)^2+b\left(b-c\right)^2+c\left(c-a\right)^2\)
Mà \(a,b,c>0\)
\(\Rightarrow a\left(a-b\right)^2+b\left(b-c\right)^2+c\left(c-a\right)^2\ge0\)
\(\Rightarrow\)\(3\left(a^2+b^2+c^2\right)\ge3\left(a^2b+b^2c+c^2a\right)\)
Lại có:
\(\left(a^2+b^2+c^2\right)^2+3\left(a^2+b^2+c^2\right)\ge6\left(a^2b+b^2c+c^2a\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b+b^2c+c^2a\right)\)<đpcm>
bài trên mk làm sai rồi, mong mọi người thông cảm và nghĩ cách khác nha
câu a (a+b+c)2 +(a+b-c)2 - 4c2= (a+b+c)2+(a+b-c+2c).(a+b-c-2c) =(a+b+c)2 +(a+b+c).(a+b-3c)=(a+b+c). (a+b+c+a+b-3c)=(a+b+c).2.(a+b-c)
câu b 4a2b2-(a2+b2-c2) = (2ab-a2-b2+c2).(2ab+a2+b2-c2)
= (c2-(a-b)2).((a+b)2-c2)
= (c-a+b).(c+a-b).(a+b-c).(a+b+c)
câu c a4+b4+c4-2a2b2+2b2c2-2a2c2-4b2c2=(a2-b2-c2)2-4b2c2=(a2-b2-c2-2bc).(a2-b2-c2+2bc)=(a2-(b+c)2).(a2-(b-c)2)=(a-b-c).(a+b+c).(a-b+c).(a+b-c)
câu d dùng pp xét giá trị riêng thay b =c (bạn tự giải ) thì đa thức này nếu coi là đa thức biến b thì đa thức A chia hết cho b-c
a,b,c bình đẳng => A chia hết cho c-a , a-b
=>A= k(a-b)(b-c)(c-a)
thay thử một bộ a,b,c bất kì => k=? (mình đang vội )
thay k tính đc vàoA= k(a-b)(b-c)(c-a)
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
1. Ta có: \(ab+bc+ca=3abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Đặt \(\hept{\begin{cases}\frac{1}{a}=m\\\frac{1}{b}=n\\\frac{1}{c}=p\end{cases}}\) khi đó \(\hept{\begin{cases}m+n+p=3\\M=2\left(m^2+n^2+p^2\right)+mnp\end{cases}}\)
Áp dụng Cauchy ta được:
\(\left(m+n-p\right)\left(m-n+p\right)\le\left(\frac{m+n-p+m-n+p}{2}\right)^2=m^2\)
\(\left(n+p-m\right)\left(n+m-p\right)\le n^2\)
\(\left(p-n+m\right)\left(p-m+n\right)\le p^2\)
\(\Rightarrow\left(m+n-p\right)\left(n+p-m\right)\left(p+m-n\right)\le mnp\)
\(\Leftrightarrow m^3+n^3+p^3+3mnp\ge m^2n+mn^2+n^2p+np^2+p^2m+pm^2\)
\(\Leftrightarrow\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-pm\right)+6mnp\ge mn\left(m-n\right)+np\left(n-p\right)+pm\left(p-m\right)\)
\(=mn\left(3-p\right)+np\left(3-m\right)+pm\left(3-n\right)\)
\(\Leftrightarrow3\left(m^2+n^2+p^2\right)-3\left(mn+np+pm\right)+6mnp\ge3\left(mn+np+pm\right)-3mnp\)
\(\Leftrightarrow3\left(m^2+n^2+p^2\right)+9mnp\ge6\left(mn+np+pm\right)\)
\(\Leftrightarrow xyz\ge\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)
\(\Rightarrow M\ge2\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)
\(=\frac{5}{3}\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)\)
\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m^2+n^2+p^2+2mn+2np+2pm\right)\)
\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m+n+p\right)^2\)
\(\ge\frac{4}{3}\cdot3+\frac{1}{3}\cdot3^2=4+3=7\)
Dấu "=" xảy ra khi: \(m=n=p=1\Leftrightarrow a=b=c=1\)
1. (a2+b2+ab)2-a2b2-b2c2-c2a2
=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2
=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2
=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)
=(a2+b2)[(a+b)2-c2]
=(a2+b2)(a+b+c)(a+b-c)
2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2
3. a(b3-c3)+b(c3-a3)+c(a3-b3)
=ab3-ac3+bc3-ba3+ca3-cb3
=a3(c-b)+b3(a-c)+c3(b-a)
=a3(c-b)-b3(c-a)+c3(b-a)
=a3(c-b)-b3(c-b+b-a)+c3(b-a)
=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)
=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)
=(a-b)(c-b)(a2+ab+2b2+bc+c2)
4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)
5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]
=2b(3a2+b2)
6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]
=(x-y-1)(x2+y2+xy-2x-y+1)
7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)
(Đúng nhớ like nhá !)
Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi