Đặt \(x+y=a;y+z=b;z+x=c\)thì P=Q có nghĩa là:
\(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)
\(\Leftrightarrow a=b=c\Leftrightarrow x+y=y+z=z+x\Leftrightarrow x=y=z\)
Cho P = \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\)
Q = (x+ y)(y+z)+(y+z)(z+x)+(z+x)(x+y)
Chứng minh rằng nếu P= Q thì x = y = z
Cho P=(x+y)^2+(y+z)^2+(x+z)^2
Q=(x+y)(y+z)(x+z)
Chứng minh: Nếu P=Q thì x=y=z
Chứng minh rằng: Nếu \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=\left(y+z-2x\right)^2+\left(z+x-2y\right)^2+\left(x+y-2z\right)^2\) thì \(x=y=z\)
chứng minh rằng
a) nếu (x-y)^2+(y-z)^2+(z-x)^2=(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2 thì x=y=z
Cho x , y , z > 0 . Chứng minh rằng : \(\frac{x^2-z^2}{y+z}+\frac{y^2-x^2}{z+x}+\frac{z^2-y^2}{x+y}\ge0\)
Chứng minh rằng:
a, nếu x+y=1 thì \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
b, nếu x,y,z khác -1 thì\(\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+z+y+1}+\frac{zx+2z+1}{zx+z+x+1}=3\)
c, Cho x,y,z đôi một khác nhau thỏa mãn\(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\) thì\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=0\)
cho x,y,z khác 0 và x+y+z=0
chứng minh rằng
\(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{x^2+z^2}{x+z}=\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\)
\(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\) ; \(Q=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Chứng minh rằng: Nếu \(P=1\) thì \(Q=0\)
1) Chứng minh rằng nếu: xyz=1 thì \(\frac{1}{1+x+xy}\)+\(\frac{1}{1+y+yz}\)+\(\frac{1}{1+z+zx}\)=1
2) Cho \(\frac{x^2}{x+y}\)+\(\frac{y^2}{y+z}\)+\(\frac{z^2}{z+x}\)=2017. Tính: \(\frac{x^2}{x+y}\)+\(\frac{z^2}{y+z}\)+\(\frac{x^2}{z+x}\)
