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\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\Rightarrow\frac{x+y+z}{xyz}=0\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(N=\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=\frac{x^3+y^3+z^3}{xyz}=\frac{3xyz}{xyz}=3\)
Đặt bài toán phụ : Chứng minh nếu \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
Thật vậy :
\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)
\(a+b=-c\)
\(b+c=-a\)
\(c+a=-b\)
\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=-3\left(-c\right)\left(-b\right)\left(-a\right)\)
\(=3abc\)
Trở lại bài toán chính :
Ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Rightarrow xy+xz+yz=0\)
\(\Rightarrow\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3\left(xy\right)\left(xz\right)\left(yz\right)=3x^2y^2z^2\)
Lại có:
\(P=\frac{xy.y^2x^2}{x^2y^2z^2}+\frac{xz.z^2.x^2}{x^2y^2z^2}+\frac{z^2.y^2.yz}{x^2y^2z^2}\)
\(=\frac{\left(xy\right)^3}{x^2y^2z^2}+\frac{\left(xz\right)^3}{x^2y^2z^2}+\frac{\left(yz\right)^3}{x^2y^2z^2}\)
\(=\frac{\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)}{x^2y^2z^2}\)
Thay \(\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3x^2y^2z^2;\)ta có:
\(P=\frac{3x^2y^2z^2}{x^2y^2z^2}\)
\(=3\)
Vậy \(P=3.\)
bạn nào cần giải ko mk giải cho
Cho mik cách giải ik :>