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\(P=x^3\left(z-y^2\right)+y^3\left(x-z^2\right)+z^3\left(y-x^2\right)+xyz\left(xyz-1\right)\)
\(P=\left(-x^3\left(y^2-z\right)\right)+xy^3-y^3z^2+yz^3-x^2z^3+x^2y^2z^2-xyz\)
\(P=\left(-x^3\left(y^2-z\right)\right)+\left(xy^3-xyz\right)-\left(y^3z^2-yz^3\right)+\left(x^2y^2z^2-x^2z^3\right)\)
\(P=\left(-x^3\left(y^2-z\right)\right)+\left(xy\left(y^2-z\right)\right)-\left(yz^2\left(y^2-z\right)\right)+\left(x^2z^2\left(y^2-z\right)\right)\)
\(P=\left(-x^3+xy-yz^2+x^2z^2\right)\left(y^2-z\right)\)
\(P=\left(\left(x^2z^2-x^3\right)-\left(yz^2-xy\right)\right)\left(y^2-z\right)\)
\(P=\left(x^2\left(z^2-x\right)-y\left(z^2-x\right)\right)\left(y^2-z\right)\)
\(P=\left(\left(x^2-y\right)\left(z^2-x\right)\right)\left(y^2-z\right)\)
\(P=\left(a.c\right).b\)
\(P=a.b.c\)
Vậy giá trị của P không phụ thuộc vào biến x;y;z (điều cần chứng minh)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
Trừ cả hai vế cho \(a^2x^2+b^2y^2+c^2z^2\), có :
\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)
\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)
\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2x^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)
\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Mà \(\hept{\begin{cases}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}ay=bx\\az=cx\\bz-cy\end{cases}}\)
\(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Vậy ...
\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=1\)
Ta có: \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)\)
Mặt khác, \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) => \(\frac{ayz+bxz+cxy}{xyz}=0\)=> ayz + bxz + cxy = 0
=> \(\frac{ayz+bxz+cxy}{abc}=0\) => \(\frac{yz}{bc}+\frac{xz}{ac}+\frac{xy}{ab}=0\)
Do đó, \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)=0\)
=> đpcm
P = ...
\(\Leftrightarrow P=x^3z-x^3y^2+y^3x-y^3z^2+z^3y-z^3x^2+x^2y^2z^2-xyz\)\(\Leftrightarrow P=\left(x^3z-x^2z^3\right)-\left(x^3y^2-x^2y^2z^2\right)+\left(xy^3-y^3z\right)+\left(yz^3-xyz\right)\)\(\Leftrightarrow P=x^2z\left(x-z^2\right)-x^2y^2\left(x-z^2\right)+y^3\left(x-z^2\right)-yz\left(x-z^2\right)\)\(\Leftrightarrow P=\left(x-z^2\right)\left(x^2z-x^2y^2+y^3-yz\right)\)
\(\Leftrightarrow P=\left(x-z^2\right)\left[\left(x^2z-x^2y^2\right)+\left(y^3-yz\right)\right]\)
\(\Leftrightarrow P=\left(x-z^2\right)\left[-x^2\left(y^2-z\right)+y\left(y^2-z\right)\right]\)
\(\Leftrightarrow P=\left(x-z\right)^2\left(y^2-z\right)\left(y-x^2\right)\)
\(\Leftrightarrow P=abc\left(đpcm\right)\)
Sửa lại
P = ...
\(\Leftrightarrow P=...\)
\(\Leftrightarrow P=...-...+\left(xy^3-y^3z^2\right)+...\)
Câu hỏi của Yến Trần - Toán lớp 8 - Học toán với OnlineMath
\(x^3\left(z-y^2\right)+y^3\left(x-z^2\right)+z^3\left(y-x^2\right)+xyz\left(xyz-1\right)\)
\(=x^3z-x^3y^2+y^3x-y^3z^2+z^3y-z^3x^2+y^2z^2x^2-xyz\)
\(=y^2z^2x^2-y^3z^2-z^3x^2+z^3y-x^3y^2+y^3x+x^3z-xyz\)
\(=y^2z^2\left(x^2-y\right)-z^3\left(x^2-y\right)-xy^2\left(x^2-y\right)+xz\left(x^2-y\right)\)
\(=\left(x^2-y\right)\left(y^2z^2-z^3-xy^2+xz\right)=\left(x^2-y\right)\left[y^2\left(z^2-x\right)-z\left(z^2-x\right)\right]\)
\(=\left(x^2-y\right)\left(z^2-x\right)\left(y^2-z\right)=a.b.c\)
Vậy P không phụ thuộc vào x,y,z