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\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(=x^4y-x^4z+y^4z-y^4x+z^4\left(x-y\right)\)
\(=xy\left(x^3-y^3\right)-z\left(x^4-y^4\right)+z^4\left(x-y\right)\)
\(=xy\left(x-y\right)\left(x^2+xy+y^2\right)-z\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)+z^4\left(x-y\right)\)
\(=\left(x-y\right)\left[xy\left(x^2+xy+y^2\right)-z\left(x^3+x^2y+xy^2+y^3\right)+z^4\right]\)
\(=\left(x-y\right)\left(x^3y+x^2y^2+xy^3-x^3z-x^2yz-xy^2z-y^3z+z^4\right)\)
\(=\left(x-y\right)\left[x^3\left(y-z\right)+x^2y\left(y-z\right)+xy^2\left(y-z\right)-z\left(y^3-z^3\right)\right]\)
\(=\left(x-y\right)\left[x^3\left(y-z\right)+x^2y\left(y-z\right)+xy^2\left(y-z\right)-z\left(y-z\right)\left(y^2+yz+z^2\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left[x^3+x^2y+xy^2-z\left(y^2+yz+z^2\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x^3+x^2y+xy^2-y^2z-yz^2-z^3\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[x^3-z^3+y\left(x^2-z^2\right)+y^2\left(x-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x^2+xz+z^2\right)+y\left(x-z\right)\left(x+z\right)+y^2\left(x-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left[x^2+xz+z^2+y\left(x+z\right)+y^2\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\frac{2\left(x^2+xz+z^2+xy+yz+y^2\right)}{2}\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\frac{x^2+2xz+z^2+x^2+xy+y^2+y^2+yz+z^2}{2}\)
\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\frac{\left(x+z\right)^2+\left(x+y\right)^2+\left(y+z\right)^2}{2}\)
\(Ta\)\(có\)\(x>y>z\Rightarrow\left(x-y\right);\left(y-z\right);\left(x-z\right)>0\)
\(\left(x+z\right)^2;\left(y+z\right)^2;\left(x+y\right)^2\ge0\)
\(\Rightarrow A>o\Rightarrow A\)\(luôn\)\(dương\)
a)Ta có: ab+ac+bc=-7 (ab+ac+bc)^2=49
nên
(ab)^2+(bc)^2+(ac)^2=49
nên a^4+b^4+c^4=(a^2+b^2+c^2)^2−2(ab)^2−2(ac)^2−2(bc^)2=98
b) (x^2+y^2+z^2)/(a^2+b^2+c^2)=
=x^2/a^2+y^2/b^2+z^2/c^2 <=>
x^2+y^2+z^2=x^2+(a^2/b^2)y^2+
+(a^2/c^2)z^2+(b^2/a^2)x^2+y^2+
+(b^2/c^2)z^2+(c^2/a^2)x^2+
+(c^2/b^2)y^2+z^2 <=>
[(b^2+c^2)/a^2]x^2+[(a^2+c^2)/b^2]y^2+
+[(a^2+b^2)/c^2]z^2 = 0 (*)
Đặt A=[(b^2+c^2)/a^2]x^2; B=[(a^2+c^2)/b^2]y^2;
và C=[(a^2+b^2)/c^2]z^2
Vì a,b,c khác 0 nên suy ra A,B,C đều không âm
Từ (*) ta có A+B+C=0
Tổng 3 số không âm bằng 0 thì cả 3 số đều phải bằng 0,tức A=B=C=0
Vì a,b,c khác 0 nên [(b^2+c^2)/c^2]>0 =>x^2=0 =>x=0
Tương tự B=C=0 =>y^2=z^2=0 => y=z=0
Vậy x^2011+y^2011+z^2011=0
Và x^2008+y^2008+z^2008=0.
ĐKXĐ: ...
a/ \(A=x-2009-4\sqrt{x-2009}+4=\left(\sqrt{x-2009}-2\right)^2\ge0\)
\(A_{min}=0\) khi \(\sqrt{x-2009}-2=0\Rightarrow x=2013\)
b/ \(\frac{1}{4}-\frac{\sqrt{x-2009}-1}{x-2009}+\frac{1}{4}-\frac{\sqrt{y-2010}-1}{y-2010}+\frac{1}{4}-\frac{\sqrt{z-2011}-1}{z-2011}=0\)
\(\Leftrightarrow\frac{x-2009-4\sqrt{x-2009}+4}{4\left(x-2009\right)}+\frac{y-2010-4\sqrt{y-2010}+4}{4\left(y-2010\right)}+\frac{z-2011-4\sqrt{z-2011}+4}{4\left(z-2011\right)}=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)
2.Câu hỏi của Lãnh Hàn Thần - Toán lớp 8 - Học toán với OnlineMath
Lời giải:
Đặt $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=t$
$\Rightarrow a=xt; b=yt; c=zt$. Ta có:
$a+b+c=xt+yt+zt=t(x+y+z)=t$
$a^2+b^2+c^2=t^2(x^2+y^2+z^2)=t^2$
$ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{t^2-t^2}{2}=0$
Ta có đpcm.
x2+y2+z2= xy+yz+zx.
=> 2x2+2y2+2z2-2xy-2yz-2zx=0
=> ( x-y)2+(y-z.)2+(z-x)2 =0
=> x=y=z=0
Thay x=y=z vào x2011+y2011+z2011=32012 ta được:
3.x2011=3.32011
=> x2011=32011
=> x=3 hoặc x = -3
Hay x=y=z=3 hoặc x=y=z=-3
1) có bn giải rồi ko giải nữa
2) \(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2011^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2012^4+\frac{1}{4}\right)}\)
Với mọi n thuộc N ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)
\(=\left[n\left(n-1\right)+\frac{1}{2}\right]\left[n\left(n+1\right)+\frac{1}{2}\right]\)
Áp dụng ta được :
\(A=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(2011.2012+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right).......\left(2012.2013+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{2012.2013+\frac{1}{2}}=\frac{1}{8100313}\)