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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=0\)
\(\Leftrightarrow ab+bc+ca=0\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2=1\)
Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
=> ab+bc+ac=0
Mà \(a+b+c=1\Leftrightarrow\left(a+b+c\right)^2=1\)
\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=1\)
\(a^2+b^2+c^2=1\)
_Kudo_
(a^2+b^2+c^2) x 2 = 2 x (a^4+b^4+c^4)
suy ra: (a+b+c)^2 x 2 = (a+b+c)^4 x 2
Mà a+b+c= 0(gt)
suy ra: 0^2 x 2=0^4 x 2
0 = 0
=)))
Vì \(0\le x,y,z\le1\) nên ta có:
\(\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\Rightarrow x^2+y^2+z^2\le x+y+z=2\)
1/ Đặt
\(\frac{a}{b^2}=x,\frac{b}{c^2}=y,\frac{c}{a^2}=z,xyz=1\)thì ta có
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow xy+yz+zx=x+y+z\)
\(\Leftrightarrow xyz-xy-yz-zx+x+y+z-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
\(\Leftrightarrow x=1;y=1;z=1\)
\(\Rightarrow\frac{a}{b^2}=1;\frac{b}{c^2}=1;\frac{c}{a^2}=1\)
\(\Leftrightarrow a=b^2;b=c^2;c=a^2\)
2/ Đặt
\(ab=x,bc=y,ca=z\) cần tính
\(P=\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\left(1+\frac{y}{x}\right)\)
\(\Rightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
Xét \(x+y+z=0\)
\(\Rightarrow P=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-1\)
Xét \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)