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3) áp dụng đẳng thức \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
<=>\(1-3xyz=1\left(1-xy-yz-zx\right)\)
<=>\(3xyz=xy+yz+zx\)
mặt khác ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx=1\)
<=>\(1+2xy+2yz+2zx=1\)
<=> \(xy+yz+zx=0\)
do đó 3xyz=0<=> \(\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
lần lượt thay x;y;z vào hệ ta có các cặp nghiệm (x;y;z)=(0;0;1),(0;1;0),(1;0;0)
do đó x^2017+y^2017+z^2017=1
\(Q=\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^3}{\left(x-y\right)\left(x-z\right)}-\frac{y^3}{\left(x-y\right)\left(y-z\right)}+\frac{z^3}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^3\left(y-z\right)-y^3\left(x-z\right)+z^3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)(1)
Ta có:
\(x^3\left(y-z\right)-y^3\left(x-z\right)+z^3\left(x-y\right)\)
\(=x^3\left(y-z\right)-y^3\left(y-z\right)-y^3\left(x-y\right)+z^3\left(x-y\right)\)
\(=\left(y-z\right)\left(x^3-y^3\right)-\left(x-y\right)\left(y^3-z^3\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(y-z\right)\left(y^2+yz+z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x^2+xy+y^2-y^2-yz-z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x^2+xy-yz-z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x+z\right)+y\left(x-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+y+z\right)=1000\left(x-y\right)\left(y-z\right)\left(x-z\right)\)(2)
Từ (1) và (2), ta có Q = 1000
Đặt \(\sqrt{x}=a\) , \(\sqrt{y}=b\) , \(\sqrt{z}=c\)
Suy ra \(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Xét tử : \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left[-\left(a-b\right)-\left(c-a\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(c^2-a^2\right)+\left(c-a\right)\left(b^2-a^2\right)=\left(a-b\right)\left(c-a\right)\left(c+a\right)+\left(c-a\right)\left(b-a\right)\left(b+a\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c+a-a-b\right)=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
Suy ra \(P=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
dễ mà bạn :))) gáy tí , sai thì thôi
\(P=\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{y^3}{\left(1+y\right)\left(1+z\right)}+\frac{z^3}{\left(1+z\right)\left(1+x\right)}\)
\(=\frac{x^3\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}+\frac{y^3\left(1+x\right)}{\left(1+y\right)\left(1+x\right)\left(1+z\right)}+\frac{z^3\left(1+y\right)}{\left(1+x\right)\left(1+z\right)\left(1+y\right)}\)
\(=\frac{x^3\left(1+z\right)+y^3\left(1+x\right)+z^3\left(1+y\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\frac{3\sqrt[3]{x^3y^3z^3\left(1+x\right)\left(1+y\right)\left(1+z\right)}}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
đến đây áp dụng BĐT phụ ( 1+a ) ( 1+b ) ( 1+c ) >= 8abc
EZ :)))
Câu hỏi của Đỗ Tuấn Linh - Toán lớp 9 - Học toán với OnlineMath
\(P=\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-x^3\left(y-z\right)-y^3\left(z-x\right)-z^3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-x^3y+x^3z-y^3z+y^3x-z^3x+z^3y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=x+y+z=2008\)